Sec 2 G1 Maths Tutor Bukit Timah | Simultaneous Equations

Accuracy note: All equations and imaginary price stories are original teaching illustrations. The 2027 published SEC syllabus is a graduating-cohort reference, not a promise of present Secondary 2 assessment content. G1 subject-level placement and programme availability must be verified with the school or centre; no grade outcome is guaranteed.

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Parents can ask: ‘What does this pair have to satisfy?’ If the child answers ‘both original equations’ and can demonstrate it without hints, the key meaning of simultaneous solution is emerging. If the pupil can only repeat a remembered elimination trick, use simpler values and revisit the equality principle.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

MOE’s 2020 Secondary Two G1 outline lists linear equations in one variable, not simultaneous two-variable systems. The full 2027 K110 graduating syllabus includes substitution, elimination and graphical methods. Treat this as next-stage preparation subject to school advice.

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

10 minutes: retrieve one-variable equation balance and ordered-pair plotting.
20 minutes: tutor models substitution from a straightforward linear condition.
25 minutes: solve differentiated systems and diagnose each student’s first wrong line.
15 minutes: compare elimination with graphical interpretation.
10 minutes: independently solve an unseen changed-number system.
10 minutes: check both original equations and assign a delayed retest.

A practical five-session bridge for parents

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

10 minutes: retrieve one-variable equation balance and ordered-pair plotting.
20 minutes: tutor models substitution from a straightforward linear condition.
25 minutes: solve differentiated systems and diagnose each student’s first wrong line.
15 minutes: compare elimination with graphical interpretation.
10 minutes: independently solve an unseen changed-number system.
10 minutes: check both original equations and assign a delayed retest.

A practical five-session bridge for parents

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

A one-minute prerequisite diagnostic
Substitution: use one equation inside another
Elimination: remove one unknown
Graphical method and intersections
More than 80 original worked clinics
Five common first-line errors
Small-group tutorial model
A parent-friendly practice schedule
FAQs
Official sources

A 60-second check before teaching two unknowns

Ask the student to solve 3x+5=20 and to rearrange y=2x+3 for x. The first gives x=5; the second gives x=(y−3)/2. Then check the meaning of the point (2,7): it satisfies y=2x+3. If solving one-variable equations or reading ordered pairs is uncertain, rebuild those prerequisite skills before adding a second equation. Simultaneous equations do not fix earlier sign or equality mistakes automatically.

Substitution: make one variable speak through the other

For x+y=9 and 2x−y=6, the first equation gives x=9−y. Put the entire expression 9−y into the second: 2(9−y)−y=6. Rearrange to 18−3y=6, giving y=4 and x=5. The answer is the ordered pair (5,4). Both original conditions now hold: 5+4=9 and 2(5)−4=6.

Brackets matter. Replacing x=9−y inside 2x does not produce 18−y; it produces 18−2y. A learner who has not mastered multiplying a negative number into a bracket needs that repair before long substitution drills. After solving, say what the first and second coordinates mean rather than treating them as two unrelated answers.

Elimination: preserve equality while removing one letter

Suppose 2x+y=11 and 2x−y=5. Add both left sides and both right sides: 4x=16, so x=4. Substitute into 2x+y=11 to get y=3. Here addition makes the opposite y terms cancel. In another system, subtraction may be easier; in a third, one or both complete equations must be multiplied first. You must multiply every term—including the right side—when making the coefficients match.

Do not over-memorise ‘always add’ or ‘always subtract’. Choose the operation that cancels the chosen variable. If the same coefficient appears with the same sign, subtraction removes it. If equal magnitudes have opposite signs, addition removes it. A compact alignment of the two complete equations helps prevent sign and transcription errors.

Graphical solutions connect back to Secondary 2 coordinates

Every linear equation in two variables describes a line. When two nonparallel lines are drawn on the same correctly scaled axes, their intersection gives a pair that satisfies both. In a school question, a plotted intersection might give an approximate result if the graph scale or drawing is imprecise; algebraic substitution can verify it exactly. Parallel distinct lines have no common point, while coincident lines share infinitely many solutions. The examples below mainly use pairs with unique intersections.

A graph should show labelled axes and a suitable scale. If a pupil confuses (x,y) order or uses the wrong gradient from an axis, the plotted solution may be misleading. Repair coordinate-table foundations rather than assuming a visual intersection is automatically correct.

86 original simultaneous-equation learning clinics

This collection includes substitution with simple first equations, direct elimination, coefficient matching, two-price word problems and graphical interpretations. All story prices are fictional. It is a selection bank, not homework to complete in one sitting. Choose a few cases that match the learner’s present stage, cover the worked method and retest with altered numbers later.

1. Substitution 1: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=-7−y. Substitute this entire expression for x in the second: 2(-7−y)−y=-5. Collect to get 2(-7)−3y=-5, so −3y=9 and y=-3. Then x=-7−(-3)=-4.

First wrong step to avoid. A learner may replace x with -7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(-3)=-7; second is 2(-4)−(-3)=-5. Both match their right sides. This verifies the ordered solution (-4, -3).

Transfer task. Keep the equation x+y=-7 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

2. Substitution 2: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-2. Collect to get 2(5)−3y=-2, so −3y=-12 and y=4. Then x=5−(4)=1.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(4)=5; second is 2(1)−(4)=-2. Both match their right sides. This verifies the ordered solution (1, 4).

Transfer task. Keep the equation x+y=5 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

3. Substitution 3: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 10. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=10. Collect to get 2(8)−3y=10, so −3y=-6 and y=2. Then x=8−(2)=6.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(2)=8; second is 2(6)−(2)=10. Both match their right sides. This verifies the ordered solution (6, 2).

Transfer task. Keep the equation x+y=8 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

4. Substitution 4: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=0. Collect to get 2(0)−3y=0, so −3y=0 and y=0. Then x=0−(0)=0.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(0)=0; second is 2(0)−(0)=0. Both match their right sides. This verifies the ordered solution (0, 0).

Transfer task. Keep the equation x+y=0 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

5. Substitution 5: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 12. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=12. Collect to get 2(3)−3y=12, so −3y=6 and y=-2. Then x=3−(-2)=5.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(-2)=3; second is 2(5)−(-2)=12. Both match their right sides. This verifies the ordered solution (5, -2).

Transfer task. Keep the equation x+y=3 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

6. Substitution 6: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = -7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=-7. Collect to get 2(4)−3y=-7, so −3y=-15 and y=5. Then x=4−(5)=-1.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(5)=4; second is 2(-1)−(5)=-7. Both match their right sides. This verifies the ordered solution (-1, 5).

Transfer task. Keep the equation x+y=4 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

7. Substitution 7: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 7 and 2x − y = 5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=7−y. Substitute this entire expression for x in the second: 2(7−y)−y=5. Collect to get 2(7)−3y=5, so −3y=-9 and y=3. Then x=7−(3)=4.

First wrong step to avoid. A learner may replace x with 7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(3)=7; second is 2(4)−(3)=5. Both match their right sides. This verifies the ordered solution (4, 3).

Transfer task. Keep the equation x+y=7 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

8. Substitution 8: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -1 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=-1−y. Substitute this entire expression for x in the second: 2(-1−y)−y=-5. Collect to get 2(-1)−3y=-5, so −3y=-3 and y=1. Then x=-1−(1)=-2.

First wrong step to avoid. A learner may replace x with -1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -2+(1)=-1; second is 2(-2)−(1)=-5. Both match their right sides. This verifies the ordered solution (-2, 1).

Transfer task. Keep the equation x+y=-1 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

9. Substitution 9: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 2 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=2−y. Substitute this entire expression for x in the second: 2(2−y)−y=7. Collect to get 2(2)−3y=7, so −3y=3 and y=-1. Then x=2−(-1)=3.

First wrong step to avoid. A learner may replace x with 2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 3+(-1)=2; second is 2(3)−(-1)=7. Both match their right sides. This verifies the ordered solution (3, -1).

Transfer task. Keep the equation x+y=2 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

10. Substitution 10: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -6 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=-6−y. Substitute this entire expression for x in the second: 2(-6−y)−y=-3. Collect to get 2(-6)−3y=-3, so −3y=9 and y=-3. Then x=-6−(-3)=-3.

First wrong step to avoid. A learner may replace x with -6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -3+(-3)=-6; second is 2(-3)−(-3)=-3. Both match their right sides. This verifies the ordered solution (-3, -3).

Transfer task. Keep the equation x+y=-6 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

11. Substitution 11: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=6−y. Substitute this entire expression for x in the second: 2(6−y)−y=0. Collect to get 2(6)−3y=0, so −3y=-12 and y=4. Then x=6−(4)=2.

First wrong step to avoid. A learner may replace x with 6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 2+(4)=6; second is 2(2)−(4)=0. Both match their right sides. This verifies the ordered solution (2, 4).

Transfer task. Keep the equation x+y=6 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

12. Substitution 12: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -2 and 2x − y = -10. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=-2−y. Substitute this entire expression for x in the second: 2(-2−y)−y=-10. Collect to get 2(-2)−3y=-10, so −3y=-6 and y=2. Then x=-2−(2)=-4.

First wrong step to avoid. A learner may replace x with -2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(2)=-2; second is 2(-4)−(2)=-10. Both match their right sides. This verifies the ordered solution (-4, 2).

Transfer task. Keep the equation x+y=-2 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

13. Substitution 13: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 1 and 2x − y = 2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=1−y. Substitute this entire expression for x in the second: 2(1−y)−y=2. Collect to get 2(1)−3y=2, so −3y=0 and y=0. Then x=1−(0)=1.

First wrong step to avoid. A learner may replace x with 1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(0)=1; second is 2(1)−(0)=2. Both match their right sides. This verifies the ordered solution (1, 0).

Transfer task. Keep the equation x+y=1 but change the coefficient of x in the other equation; the pair generally changes. Ask which line in the substitution process would change and why, without immediately solving every step.

14. Substitution 14: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = 14. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=14. Collect to get 2(4)−3y=14, so −3y=6 and y=-2. Then x=4−(-2)=6.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(-2)=4; second is 2(6)−(-2)=14. Both match their right sides. This verifies the ordered solution (6, -2).

15. Substitution 15: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-5. Collect to get 2(5)−3y=-5, so −3y=-15 and y=5. Then x=5−(5)=0.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(5)=5; second is 2(0)−(5)=-5. Both match their right sides. This verifies the ordered solution (0, 5).

16. Substitution 16: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=7. Collect to get 2(8)−3y=7, so −3y=-9 and y=3. Then x=8−(3)=5.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(3)=8; second is 2(5)−(3)=7. Both match their right sides. This verifies the ordered solution (5, 3).

17. Substitution 17: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=-3. Collect to get 2(0)−3y=-3, so −3y=-3 and y=1. Then x=0−(1)=-1.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(1)=0; second is 2(-1)−(1)=-3. Both match their right sides. This verifies the ordered solution (-1, 1).

18. Substitution 18: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 9. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=9. Collect to get 2(3)−3y=9, so −3y=3 and y=-1. Then x=3−(-1)=4.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(-1)=3; second is 2(4)−(-1)=9. Both match their right sides. This verifies the ordered solution (4, -1).

19. Elimination 1: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and x − y = -1. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=-8, so x=-4. Substitute into the first equation: 1(-4)+1y=-7, meaning 1y=-3. Hence y=-3.

First wrong step to avoid. If the pupil subtracts these equations instead of adding, they will eliminate x and must follow a different valid route. Adding the right sides but not the left sides does not preserve the system.

Verify in both original equations. Equation one: -4+(-3)=-7. Equation two: -4−(-3)=-1. Both checks hold even when a found value is negative.

Transfer task. Explain how a change in the sign of the second y-term alters whether addition or subtraction is the easiest elimination move. Solve a changed system by the other valid method and compare results.

20. Elimination 2: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = 10 and 2x − 2y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=12, so x=3. Substitute into the first equation: 2(3)+2y=10, meaning 2y=4. Hence y=2.

Verify in both original equations. Equation one: 6+(4)=10. Equation two: 6−(4)=2. Both checks hold even when a found value is negative.

21. Elimination 3: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = -9 and 3x − 3y = 3. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=-6, so x=-1. Substitute into the first equation: 3(-1)+3y=-9, meaning 3y=-6. Hence y=-2.

Verify in both original equations. Equation one: -3+(-6)=-9. Equation two: -3−(-6)=3. Both checks hold even when a found value is negative.

22. Elimination 4: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 18 and x − 4y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=12, so x=6. Substitute into the first equation: 1(6)+4y=18, meaning 4y=12. Hence y=3.

Verify in both original equations. Equation one: 6+(12)=18. Equation two: 6−(12)=-6. Both checks hold even when a found value is negative.

23. Elimination 5: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 3 and 2x − y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=8, so x=2. Substitute into the first equation: 2(2)+1y=3, meaning 1y=-1. Hence y=-1.

Verify in both original equations. Equation one: 4+(-1)=3. Equation two: 4−(-1)=5. Both checks hold even when a found value is negative.

24. Elimination 6: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 2 and 3x − 2y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=-12, so x=-2. Substitute into the first equation: 3(-2)+2y=2, meaning 2y=8. Hence y=4.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

25. Elimination 7: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = 5 and x − 3y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=10, so x=5. Substitute into the first equation: 1(5)+3y=5, meaning 3y=0. Hence y=0.

Verify in both original equations. Equation one: 5+(0)=5. Equation two: 5−(0)=5. Both checks hold even when a found value is negative.

26. Elimination 8: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 22 and 2x − 4y = -18. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=4, so x=1. Substitute into the first equation: 2(1)+4y=22, meaning 4y=20. Hence y=5.

Verify in both original equations. Equation one: 2+(20)=22. Equation two: 2−(20)=-18. Both checks hold even when a found value is negative.

27. Elimination 9: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + y = -8 and 3x − y = -10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 6x=-18, so x=-3. Substitute into the first equation: 3(-3)+1y=-8, meaning 1y=1. Hence y=1.

Verify in both original equations. Equation one: -9+(1)=-8. Equation two: -9−(1)=-10. Both checks hold even when a found value is negative.

28. Elimination 10: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 2y = -2 and x − 2y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 2x=8, so x=4. Substitute into the first equation: 1(4)+2y=-2, meaning 2y=-6. Hence y=-3.

Verify in both original equations. Equation one: 4+(-6)=-2. Equation two: 4−(-6)=10. Both checks hold even when a found value is negative.

29. Elimination 11: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 3y = 6 and 2x − 3y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 4x=0, so x=0. Substitute into the first equation: 2(0)+3y=6, meaning 3y=6. Hence y=2.

Verify in both original equations. Equation one: 0+(6)=6. Equation two: 0−(6)=-6. Both checks hold even when a found value is negative.

30. Elimination 12: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 4y = -20 and 3x − 4y = -4. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 6x=-24, so x=-4. Substitute into the first equation: 3(-4)+4y=-20, meaning 4y=-8. Hence y=-2.

Verify in both original equations. Equation one: -12+(-8)=-20. Equation two: -12−(-8)=-4. Both checks hold even when a found value is negative.

31. Elimination 13: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and x − y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=6, so x=3. Substitute into the first equation: 1(3)+1y=6, meaning 1y=3. Hence y=3.

Verify in both original equations. Equation one: 3+(3)=6. Equation two: 3−(3)=0. Both checks hold even when a found value is negative.

32. Elimination 14: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = -4 and 2x − 2y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=-4, so x=-1. Substitute into the first equation: 2(-1)+2y=-4, meaning 2y=-2. Hence y=-1.

Verify in both original equations. Equation one: -2+(-2)=-4. Equation two: -2−(-2)=0. Both checks hold even when a found value is negative.

33. Elimination 15: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = 30 and 3x − 3y = 6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=36, so x=6. Substitute into the first equation: 3(6)+3y=30, meaning 3y=12. Hence y=4.

Verify in both original equations. Equation one: 18+(12)=30. Equation two: 18−(12)=6. Both checks hold even when a found value is negative.

34. Elimination 16: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 2 and x − 4y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=4, so x=2. Substitute into the first equation: 1(2)+4y=2, meaning 4y=0. Hence y=0.

Verify in both original equations. Equation one: 2+(0)=2. Equation two: 2−(0)=2. Both checks hold even when a found value is negative.

35. Elimination 17: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 1 and 2x − y = -9. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=-8, so x=-2. Substitute into the first equation: 2(-2)+1y=1, meaning 1y=5. Hence y=5.

Verify in both original equations. Equation one: -4+(5)=1. Equation two: -4−(5)=-9. Both checks hold even when a found value is negative.

36. Elimination 18: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 17 and 3x − 2y = 13. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=30, so x=5. Substitute into the first equation: 3(5)+2y=17, meaning 2y=2. Hence y=1.

Verify in both original equations. Equation one: 15+(2)=17. Equation two: 15−(2)=13. Both checks hold even when a found value is negative.

37. Elimination 19: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = -8 and x − 3y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=2, so x=1. Substitute into the first equation: 1(1)+3y=-8, meaning 3y=-9. Hence y=-3.

Verify in both original equations. Equation one: 1+(-9)=-8. Equation two: 1−(-9)=10. Both checks hold even when a found value is negative.

38. Elimination 20: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 2 and 2x − 4y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=-12, so x=-3. Substitute into the first equation: 2(-3)+4y=2, meaning 4y=8. Hence y=2.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

39. Elimination with multipliers 1: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + y = -9 and 2x + y = -13. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 1. Both now contain 2x. Subtract the second resulting equation from the first: (2−1)y = -18−-13. This is 1y=-5, giving y=-5. Substitution into the original first equation gives 1x+1(-5)=-9, so x=-4.

First wrong step to avoid. The pupil may multiply only one term of an equation instead of every term including the right side. Another common mistake is reversing one subtraction but not the other.

Verify in both original equations. The original first equation gives -4+(-5)=-9; the second gives -8+(-5)=-13. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 1 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

40. Elimination with multipliers 2: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 3y = -2 and 3x + 4y = -3. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 2. Both now contain 6x. Subtract the second resulting equation from the first: (9−8)y = -6−-6. This is 1y=0, giving y=0. Substitution into the original first equation gives 2x+3(0)=-2, so x=-1.

Verify in both original equations. The original first equation gives -2+(0)=-2; the second gives -3+(0)=-3. Both must be satisfied; a correct value in only one equation is not sufficient.

41. Elimination with multipliers 3: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 5y = 31 and 4x + 7y = 43. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 3. Both now contain 12x. Subtract the second resulting equation from the first: (20−21)y = 124−129. This is -1y=-5, giving y=5. Substitution into the original first equation gives 3x+5(5)=31, so x=2.

Verify in both original equations. The original first equation gives 6+(25)=31; the second gives 8+(35)=43. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -1 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

42. Elimination with multipliers 4: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 2y = 3 and 5x + 3y = 22. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 1. Both now contain 5x. Subtract the second resulting equation from the first: (10−3)y = 15−22. This is 7y=-7, giving y=-1. Substitution into the original first equation gives 1x+2(-1)=3, so x=5.

Verify in both original equations. The original first equation gives 5+(-2)=3; the second gives 25+(-3)=22. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 7 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

43. Elimination with multipliers 5: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 4y = 10 and 2x + 6y = 18. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 2. Both now contain 4x. Subtract the second resulting equation from the first: (8−12)y = 20−36. This is -4y=-16, giving y=4. Substitution into the original first equation gives 2x+4(4)=10, so x=-3.

Verify in both original equations. The original first equation gives -6+(16)=10; the second gives -6+(24)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -4 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

44. Elimination with multipliers 6: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + y = -2 and 3x + 2y = -4. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 3. Both now contain 9x. Subtract the second resulting equation from the first: (3−6)y = -6−-12. This is -3y=6, giving y=-2. Substitution into the original first equation gives 3x+1(-2)=-2, so x=0.

Verify in both original equations. The original first equation gives 0+(-2)=-2; the second gives 0+(-4)=-4. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -3 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

45. Elimination with multipliers 7: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 3y = 12 and 4x + 5y = 27. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 1. Both now contain 4x. Subtract the second resulting equation from the first: (12−5)y = 48−27. This is 7y=21, giving y=3. Substitution into the original first equation gives 1x+3(3)=12, so x=3.

Verify in both original equations. The original first equation gives 3+(9)=12; the second gives 12+(15)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

46. Elimination with multipliers 8: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 5y = -3 and 5x + y = 27. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 2. Both now contain 10x. Subtract the second resulting equation from the first: (25−2)y = -15−54. This is 23y=-69, giving y=-3. Substitution into the original first equation gives 2x+5(-3)=-3, so x=6.

Verify in both original equations. The original first equation gives 12+(-15)=-3; the second gives 30+(-3)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 23 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

47. Elimination with multipliers 9: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 2y = -2 and 2x + 4y = 4. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 3. Both now contain 6x. Subtract the second resulting equation from the first: (4−12)y = -4−12. This is -8y=-16, giving y=2. Substitution into the original first equation gives 3x+2(2)=-2, so x=-2.

Verify in both original equations. The original first equation gives -6+(4)=-2; the second gives -4+(8)=4. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -8 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

48. Elimination with multipliers 10: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 4y = -15 and 3x + 7y = -25. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 1. Both now contain 3x. Subtract the second resulting equation from the first: (12−7)y = -45−-25. This is 5y=-20, giving y=-4. Substitution into the original first equation gives 1x+4(-4)=-15, so x=1.

Verify in both original equations. The original first equation gives 1+(-16)=-15; the second gives 3+(-28)=-25. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 5 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

49. Elimination with multipliers 11: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + y = 9 and 4x + 3y = 19. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 2. Both now contain 8x. Subtract the second resulting equation from the first: (4−6)y = 36−38. This is -2y=-2, giving y=1. Substitution into the original first equation gives 2x+1(1)=9, so x=4.

Verify in both original equations. The original first equation gives 8+(1)=9; the second gives 16+(3)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -2 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

50. Elimination with multipliers 12: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 3y = -27 and 5x + 6y = -50. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 3. Both now contain 15x. Subtract the second resulting equation from the first: (15−18)y = -135−-150. This is -3y=15, giving y=-5. Substitution into the original first equation gives 3x+3(-5)=-27, so x=-4.

Verify in both original equations. The original first equation gives -12+(-15)=-27; the second gives -20+(-30)=-50. Both must be satisfied; a correct value in only one equation is not sufficient.

51. Elimination with multipliers 13: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 5y = -1 and 2x + 2y = -2. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 1. Both now contain 2x. Subtract the second resulting equation from the first: (10−2)y = -2−-2. This is 8y=0, giving y=0. Substitution into the original first equation gives 1x+5(0)=-1, so x=-1.

Verify in both original equations. The original first equation gives -1+(0)=-1; the second gives -2+(0)=-2. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 8 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

52. Elimination with multipliers 14: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 2y = 14 and 3x + 5y = 31. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 2. Both now contain 6x. Subtract the second resulting equation from the first: (6−10)y = 42−62. This is -4y=-20, giving y=5. Substitution into the original first equation gives 2x+2(5)=14, so x=2.

Verify in both original equations. The original first equation gives 4+(10)=14; the second gives 6+(25)=31. Both must be satisfied; a correct value in only one equation is not sufficient.

53. Elimination with multipliers 15: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 4y = 11 and 4x + y = 19. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 3. Both now contain 12x. Subtract the second resulting equation from the first: (16−3)y = 44−57. This is 13y=-13, giving y=-1. Substitution into the original first equation gives 3x+4(-1)=11, so x=5.

Verify in both original equations. The original first equation gives 15+(-4)=11; the second gives 20+(-1)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 13 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

54. Elimination with multipliers 16: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + y = 1 and 5x + 4y = 1. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 1. Both now contain 5x. Subtract the second resulting equation from the first: (5−4)y = 5−1. This is 1y=4, giving y=4. Substitution into the original first equation gives 1x+1(4)=1, so x=-3.

Verify in both original equations. The original first equation gives -3+(4)=1; the second gives -15+(16)=1. Both must be satisfied; a correct value in only one equation is not sufficient.

55. Elimination with multipliers 17: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 3y = -6 and 2x + 7y = -14. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 2. Both now contain 4x. Subtract the second resulting equation from the first: (6−14)y = -12−-28. This is -8y=16, giving y=-2. Substitution into the original first equation gives 2x+3(-2)=-6, so x=0.

Verify in both original equations. The original first equation gives 0+(-6)=-6; the second gives 0+(-14)=-14. Both must be satisfied; a correct value in only one equation is not sufficient.

56. Elimination with multipliers 18: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 5y = 24 and 3x + 3y = 18. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 3. Both now contain 9x. Subtract the second resulting equation from the first: (15−9)y = 72−54. This is 6y=18, giving y=3. Substitution into the original first equation gives 3x+5(3)=24, so x=3.

Verify in both original equations. The original first equation gives 9+(15)=24; the second gives 9+(9)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 6 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

57. Elimination with multipliers 19: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 2y = 0 and 4x + 6y = 6. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 1. Both now contain 4x. Subtract the second resulting equation from the first: (8−6)y = 0−6. This is 2y=-6, giving y=-3. Substitution into the original first equation gives 1x+2(-3)=0, so x=6.

Verify in both original equations. The original first equation gives 6+(-6)=0; the second gives 24+(-18)=6. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 2 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

58. Elimination with multipliers 20: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 4y = 4 and 5x + 2y = -6. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 2. Both now contain 10x. Subtract the second resulting equation from the first: (20−4)y = 20−-12. This is 16y=32, giving y=2. Substitution into the original first equation gives 2x+4(2)=4, so x=-2.

Verify in both original equations. The original first equation gives -4+(8)=4; the second gives -10+(4)=-6. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 16 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

59. Elimination with multipliers 21: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + y = -1 and 2x + 5y = -18. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 2 and the second by 3. Both now contain 6x. Subtract the second resulting equation from the first: (2−15)y = -2−-54. This is -13y=52, giving y=-4. Substitution into the original first equation gives 3x+1(-4)=-1, so x=1.

Verify in both original equations. The original first equation gives 3+(-4)=-1; the second gives 2+(-20)=-18. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -13 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

60. Elimination with multipliers 22: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system x + 3y = 7 and 3x + y = 13. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 3 and the second by 1. Both now contain 3x. Subtract the second resulting equation from the first: (9−1)y = 21−13. This is 8y=8, giving y=1. Substitution into the original first equation gives 1x+3(1)=7, so x=4.

Verify in both original equations. The original first equation gives 4+(3)=7; the second gives 12+(1)=13. Both must be satisfied; a correct value in only one equation is not sufficient.

61. Elimination with multipliers 23: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 2x + 5y = -33 and 4x + 4y = -36. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 4 and the second by 2. Both now contain 8x. Subtract the second resulting equation from the first: (20−8)y = -132−-72. This is 12y=-60, giving y=-5. Substitution into the original first equation gives 2x+5(-5)=-33, so x=-4.

Verify in both original equations. The original first equation gives -8+(-25)=-33; the second gives -16+(-20)=-36. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient 12 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

62. Elimination with multipliers 24: align x-coefficients

Did you know? A simultaneous solution must satisfy two conditions at once. Solve the two-equation system 3x + 2y = -3 and 5x + 7y = -5. Choose elimination rather than guessing individual variable values.

Worked method. Multiply the first equation by 5 and the second by 3. Both now contain 15x. Subtract the second resulting equation from the first: (10−21)y = -15−-15. This is -11y=0, giving y=0. Substitution into the original first equation gives 3x+2(0)=-3, so x=-1.

Verify in both original equations. The original first equation gives -3+(0)=-3; the second gives -5+(0)=-5. Both must be satisfied; a correct value in only one equation is not sufficient.

Transfer task. The determinant-like coefficient -11 is nonzero, so this particular pair of lines has a unique intersection. Ask the learner to explain the meaning of this nonzero elimination coefficient without requiring unfamiliar terminology.

63. Fictional purchase model 1: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 2 notebook(s) and 1 pen(s) cost $6. Another purchase of 1 notebook(s) and 3 pen(s) costs $8. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 2p+1q=6 and 1p+3q=8. Multiply the first by 1 and second by 2, then subtract: (1−6)q=-10. Hence q=$2. Substitute into 2p+1(2)=6 to obtain p=$2.

First wrong step to avoid. A child may add both totals and divide by total number of objects even though a pen and notebook need not have equal prices. One combined ‘average price’ does not reveal two individual rates.

Verify in both original equations. At these prices, the first basket costs 2×2+1×2=$6, and the second 1×2+3×2=$8. The simultaneous system reconstructs both supplied totals.

Transfer task. Change the second basket’s number of pens and its total consistently. Ask what information is needed to continue identifying both prices. All prices here are hypothetical, not actual stationery offers.

64. Fictional purchase model 2: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 1 notebook(s) and 2 pen(s) cost $13. Another purchase of 3 notebook(s) and 1 pen(s) costs $14. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 1p+2q=13 and 3p+1q=14. Multiply the first by 3 and second by 1, then subtract: (6−1)q=25. Hence q=$5. Substitute into 1p+2(5)=13 to obtain p=$3.

Verify in both original equations. At these prices, the first basket costs 1×3+2×5=$13, and the second 3×3+1×5=$14. The simultaneous system reconstructs both supplied totals.

65. Fictional purchase model 3: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 3 notebook(s) and 2 pen(s) cost $28. Another purchase of 1 notebook(s) and 4 pen(s) costs $36. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 3p+2q=28 and 1p+4q=36. Multiply the first by 1 and second by 3, then subtract: (2−12)q=-80. Hence q=$8. Substitute into 3p+2(8)=28 to obtain p=$4.

Verify in both original equations. At these prices, the first basket costs 3×4+2×8=$28, and the second 1×4+4×8=$36. The simultaneous system reconstructs both supplied totals.

66. Fictional purchase model 4: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 2 notebook(s) and 3 pen(s) cost $22. Another purchase of 4 notebook(s) and 1 pen(s) costs $24. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 2p+3q=22 and 4p+1q=24. Multiply the first by 4 and second by 2, then subtract: (12−2)q=40. Hence q=$4. Substitute into 2p+3(4)=22 to obtain p=$5.

Verify in both original equations. At these prices, the first basket costs 2×5+3×4=$22, and the second 4×5+1×4=$24. The simultaneous system reconstructs both supplied totals.

67. Fictional purchase model 5: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 1 notebook(s) and 3 pen(s) cost $27. Another purchase of 2 notebook(s) and 1 pen(s) costs $19. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 1p+3q=27 and 2p+1q=19. Multiply the first by 2 and second by 1, then subtract: (6−1)q=35. Hence q=$7. Substitute into 1p+3(7)=27 to obtain p=$6.

Verify in both original equations. At these prices, the first basket costs 1×6+3×7=$27, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

68. Fictional purchase model 6: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 2 notebook(s) and 4 pen(s) cost $26. Another purchase of 1 notebook(s) and 3 pen(s) costs $16. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 2p+4q=26 and 1p+3q=16. Multiply the first by 1 and second by 2, then subtract: (4−6)q=-6. Hence q=$3. Substitute into 2p+4(3)=26 to obtain p=$7.

Verify in both original equations. At these prices, the first basket costs 2×7+4×3=$26, and the second 1×7+3×3=$16. The simultaneous system reconstructs both supplied totals.

69. Fictional purchase model 7: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 3 notebook(s) and 1 pen(s) cost $30. Another purchase of 2 notebook(s) and 5 pen(s) costs $46. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 3p+1q=30 and 2p+5q=46. Multiply the first by 2 and second by 3, then subtract: (2−15)q=-78. Hence q=$6. Substitute into 3p+1(6)=30 to obtain p=$8.

Verify in both original equations. At these prices, the first basket costs 3×8+1×6=$30, and the second 2×8+5×6=$46. The simultaneous system reconstructs both supplied totals.

70. Fictional purchase model 8: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 1 notebook(s) and 4 pen(s) cost $10. Another purchase of 3 notebook(s) and 2 pen(s) costs $10. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 1p+4q=10 and 3p+2q=10. Multiply the first by 3 and second by 1, then subtract: (12−2)q=20. Hence q=$2. Substitute into 1p+4(2)=10 to obtain p=$2.

Verify in both original equations. At these prices, the first basket costs 1×2+4×2=$10, and the second 3×2+2×2=$10. The simultaneous system reconstructs both supplied totals.

71. Fictional purchase model 9: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 4 notebook(s) and 1 pen(s) cost $17. Another purchase of 2 notebook(s) and 3 pen(s) costs $21. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 4p+1q=17 and 2p+3q=21. Multiply the first by 2 and second by 4, then subtract: (2−12)q=-50. Hence q=$5. Substitute into 4p+1(5)=17 to obtain p=$3.

Verify in both original equations. At these prices, the first basket costs 4×3+1×5=$17, and the second 2×3+3×5=$21. The simultaneous system reconstructs both supplied totals.

72. Fictional purchase model 10: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 2 notebook(s) and 5 pen(s) cost $48. Another purchase of 3 notebook(s) and 1 pen(s) costs $20. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 2p+5q=48 and 3p+1q=20. Multiply the first by 3 and second by 2, then subtract: (15−2)q=104. Hence q=$8. Substitute into 2p+5(8)=48 to obtain p=$4.

Verify in both original equations. At these prices, the first basket costs 2×4+5×8=$48, and the second 3×4+1×8=$20. The simultaneous system reconstructs both supplied totals.

73. Fictional purchase model 11: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 3 notebook(s) and 4 pen(s) cost $31. Another purchase of 1 notebook(s) and 2 pen(s) costs $13. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 3p+4q=31 and 1p+2q=13. Multiply the first by 1 and second by 3, then subtract: (4−6)q=-8. Hence q=$4. Substitute into 3p+4(4)=31 to obtain p=$5.

Verify in both original equations. At these prices, the first basket costs 3×5+4×4=$31, and the second 1×5+2×4=$13. The simultaneous system reconstructs both supplied totals.

74. Fictional purchase model 12: two unknown prices

Did you know? A simultaneous solution must satisfy two conditions at once. At a made-up stationery stall, 4 notebook(s) and 3 pen(s) cost $45. Another purchase of 2 notebook(s) and 1 pen(s) costs $19. Find the fixed price of each item, assuming identical unit prices and no discounts.

Worked method. Let p be notebook price and q be pen price. Form 4p+3q=45 and 2p+1q=19. Multiply the first by 2 and second by 4, then subtract: (6−4)q=14. Hence q=$7. Substitute into 4p+3(7)=45 to obtain p=$6.

Verify in both original equations. At these prices, the first basket costs 4×6+3×7=$45, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

75. Graphical interpretation 1: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=x − 1 and y=2x + 2. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 1x+(-1)=2x+(2). Thus (-1)x=3, giving x=-3. Substitute in either rule to get y=-4. The graphs intersect at (-3, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: 1(-3)+(-1)=-4. Second: 2(-3)+(2)=-4. The lines have distinct slopes 1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

76. Graphical interpretation 2: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=x + 3 and y=−x − 1. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 1x+(3)=-1x+(-1). Thus (2)x=-4, giving x=-2. Substitute in either rule to get y=1. The graphs intersect at (-2, 1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 1(-2)+(3)=1. Second: -1(-2)+(-1)=1. The lines have distinct slopes 1 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

77. Graphical interpretation 3: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=2x + 8 and y=-2x + 4. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 2x+(8)=-2x+(4). Thus (4)x=-4, giving x=-1. Substitute in either rule to get y=6. The graphs intersect at (-1, 6).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 2(-1)+(8)=6. Second: -2(-1)+(4)=6. The lines have distinct slopes 2 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

78. Graphical interpretation 4: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=3x + 0 and y=x + 0. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 3x+(0)=1x+(0). Thus (2)x=0, giving x=0. Substitute in either rule to get y=0. The graphs intersect at (0, 0).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=0: 3(0)+(0)=0. Second: 1(0)+(0)=0. The lines have distinct slopes 3 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

79. Graphical interpretation 5: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=−x + 6 and y=2x + 3. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: -1x+(6)=2x+(3). Thus (-3)x=-3, giving x=1. Substitute in either rule to get y=5. The graphs intersect at (1, 5).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=1: -1(1)+(6)=5. Second: 2(1)+(3)=5. The lines have distinct slopes -1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

80. Graphical interpretation 6: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=-2x + 3 and y=x − 3. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: -2x+(3)=1x+(-3). Thus (-3)x=-6, giving x=2. Substitute in either rule to get y=-1. The graphs intersect at (2, -1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=2: -2(2)+(3)=-1. Second: 1(2)+(-3)=-1. The lines have distinct slopes -2 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

81. Graphical interpretation 7: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=2x − 2 and y=3x − 5. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 2x+(-2)=3x+(-5). Thus (-1)x=-3, giving x=3. Substitute in either rule to get y=4. The graphs intersect at (3, 4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=3: 2(3)+(-2)=4. Second: 3(3)+(-5)=4. The lines have distinct slopes 2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

82. Graphical interpretation 8: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=−x + 2 and y=-3x + 10. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: -1x+(2)=-3x+(10). Thus (2)x=8, giving x=4. Substitute in either rule to get y=-2. The graphs intersect at (4, -2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=4: -1(4)+(2)=-2. Second: -3(4)+(10)=-2. The lines have distinct slopes -1 and -3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

83. Graphical interpretation 9: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=x − 2 and y=4x − 17. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 1x+(-2)=4x+(-17). Thus (-3)x=-15, giving x=5. Substitute in either rule to get y=3. The graphs intersect at (5, 3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=5: 1(5)+(-2)=3. Second: 4(5)+(-17)=3. The lines have distinct slopes 1 and 4, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

84. Graphical interpretation 10: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=-2x − 9 and y=3x + 6. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: -2x+(-9)=3x+(6). Thus (-5)x=15, giving x=-3. Substitute in either rule to get y=-3. The graphs intersect at (-3, -3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: -2(-3)+(-9)=-3. Second: 3(-3)+(6)=-3. The lines have distinct slopes -2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

85. Graphical interpretation 11: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=3x + 8 and y=-2x − 2. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 3x+(8)=-2x+(-2). Thus (5)x=-10, giving x=-2. Substitute in either rule to get y=2. The graphs intersect at (-2, 2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 3(-2)+(8)=2. Second: -2(-2)+(-2)=2. The lines have distinct slopes 3 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

86. Graphical interpretation 12: intersection of two lines

Did you know? A simultaneous solution must satisfy two conditions at once. Two graphs obey y=4x + 0 and y=−x − 5. Find the common coordinate algebraically and explain what it means visually.

Worked method. Set the right sides equal: 4x+(0)=-1x+(-5). Thus (5)x=-5, giving x=-1. Substitute in either rule to get y=-4. The graphs intersect at (-1, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 4(-1)+(0)=-4. Second: -1(-1)+(-5)=-4. The lines have distinct slopes 4 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

Five different errors that may look like one wrong answer

Only one equation is satisfied

A pupil finds values that make 2x+y=11 true but never checks 2x−y=5. The pair may be one of many points on the first line, not the common intersection. Always substitute the final x and y into both original conditions, with each right side reconstructed separately.

Brackets are lost during substitution

If x=8−y and a later equation contains 3x, the replacement is 3(8−y)=24−3y. Writing 24−y fails to multiply the full expression. Use visible brackets until the expansion is complete, then practise changed coefficients to ensure the habit transfers.

Only one side is multiplied in elimination

Changing 2x+3y=12 into 4x+6y=12 is invalid: the right side must also double to 24. The student should speak the reason aloud: multiplying both sides preserves equality and therefore the same solution set. Check a known solution before and after scaling an equation.

A subtraction sign is reversed inconsistently

When subtracting an entire equation, every term—including a negative y term and the right-hand number—changes sign. A useful written layout places brackets around the second line before subtraction. The pupil must not negate one coefficient while forgetting another.

A graph’s y-intercept is confused with intersection

Each line has its own y-intercept. The simultaneous solution is where the two lines cross each other, which may be far from the vertical axis. Ask which point belongs to both rules and test its coordinates algebraically. A correctly plotted line is not enough if the wrong graph feature is read.

A three-pupil tutorial can support different readiness levels

The immutable eduKateSG Clementi Maths tutorial reference describes 90-minute classes for three pupils near Sixth Avenue MRT. A suitable small-group lesson can give one pupil a one-variable balancing repair, another a substitution exercise and a third an elimination-plus-graph extension. This is a teaching model, not evidence of current class availability or eligibility for a new subject level.

10 minutes: retrieve one-variable equation balance and ordered-pair plotting.
20 minutes: tutor models substitution from a straightforward linear condition.
25 minutes: solve differentiated systems and diagnose each student’s first wrong line.
15 minutes: compare elimination with graphical interpretation.
10 minutes: independently solve an unseen changed-number system.
10 minutes: check both original equations and assign a delayed retest.

A practical five-session bridge for parents

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

A one-minute prerequisite diagnostic
Substitution: use one equation inside another
Elimination: remove one unknown
Graphical method and intersections
More than 80 original worked clinics
Five common first-line errors
Small-group tutorial model
A parent-friendly practice schedule
FAQs
Official sources

A 60-second check before teaching two unknowns

Substitution: make one variable speak through the other

Elimination: preserve equality while removing one letter

Graphical solutions connect back to Secondary 2 coordinates

86 original simultaneous-equation learning clinics

1. Substitution 1: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(-3)=-7; second is 2(-4)−(-3)=-5. Both match their right sides. This verifies the ordered solution (-4, -3).

2. Substitution 2: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-2. Collect to get 2(5)−3y=-2, so −3y=-12 and y=4. Then x=5−(4)=1.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(4)=5; second is 2(1)−(4)=-2. Both match their right sides. This verifies the ordered solution (1, 4).

3. Substitution 3: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 10. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=10. Collect to get 2(8)−3y=10, so −3y=-6 and y=2. Then x=8−(2)=6.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(2)=8; second is 2(6)−(2)=10. Both match their right sides. This verifies the ordered solution (6, 2).

4. Substitution 4: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=0. Collect to get 2(0)−3y=0, so −3y=0 and y=0. Then x=0−(0)=0.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(0)=0; second is 2(0)−(0)=0. Both match their right sides. This verifies the ordered solution (0, 0).

5. Substitution 5: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 12. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=12. Collect to get 2(3)−3y=12, so −3y=6 and y=-2. Then x=3−(-2)=5.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(-2)=3; second is 2(5)−(-2)=12. Both match their right sides. This verifies the ordered solution (5, -2).

6. Substitution 6: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = -7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=-7. Collect to get 2(4)−3y=-7, so −3y=-15 and y=5. Then x=4−(5)=-1.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(5)=4; second is 2(-1)−(5)=-7. Both match their right sides. This verifies the ordered solution (-1, 5).

7. Substitution 7: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 7 and 2x − y = 5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=7−y. Substitute this entire expression for x in the second: 2(7−y)−y=5. Collect to get 2(7)−3y=5, so −3y=-9 and y=3. Then x=7−(3)=4.

First wrong step to avoid. A learner may replace x with 7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(3)=7; second is 2(4)−(3)=5. Both match their right sides. This verifies the ordered solution (4, 3).

8. Substitution 8: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -1 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -2+(1)=-1; second is 2(-2)−(1)=-5. Both match their right sides. This verifies the ordered solution (-2, 1).

9. Substitution 9: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 2 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=2−y. Substitute this entire expression for x in the second: 2(2−y)−y=7. Collect to get 2(2)−3y=7, so −3y=3 and y=-1. Then x=2−(-1)=3.

First wrong step to avoid. A learner may replace x with 2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 3+(-1)=2; second is 2(3)−(-1)=7. Both match their right sides. This verifies the ordered solution (3, -1).

10. Substitution 10: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -6 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -3+(-3)=-6; second is 2(-3)−(-3)=-3. Both match their right sides. This verifies the ordered solution (-3, -3).

11. Substitution 11: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=6−y. Substitute this entire expression for x in the second: 2(6−y)−y=0. Collect to get 2(6)−3y=0, so −3y=-12 and y=4. Then x=6−(4)=2.

First wrong step to avoid. A learner may replace x with 6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 2+(4)=6; second is 2(2)−(4)=0. Both match their right sides. This verifies the ordered solution (2, 4).

12. Substitution 12: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -2 and 2x − y = -10. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(2)=-2; second is 2(-4)−(2)=-10. Both match their right sides. This verifies the ordered solution (-4, 2).

13. Substitution 13: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 1 and 2x − y = 2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=1−y. Substitute this entire expression for x in the second: 2(1−y)−y=2. Collect to get 2(1)−3y=2, so −3y=0 and y=0. Then x=1−(0)=1.

First wrong step to avoid. A learner may replace x with 1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(0)=1; second is 2(1)−(0)=2. Both match their right sides. This verifies the ordered solution (1, 0).

14. Substitution 14: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = 14. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=14. Collect to get 2(4)−3y=14, so −3y=6 and y=-2. Then x=4−(-2)=6.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(-2)=4; second is 2(6)−(-2)=14. Both match their right sides. This verifies the ordered solution (6, -2).

15. Substitution 15: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-5. Collect to get 2(5)−3y=-5, so −3y=-15 and y=5. Then x=5−(5)=0.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(5)=5; second is 2(0)−(5)=-5. Both match their right sides. This verifies the ordered solution (0, 5).

16. Substitution 16: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=7. Collect to get 2(8)−3y=7, so −3y=-9 and y=3. Then x=8−(3)=5.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(3)=8; second is 2(5)−(3)=7. Both match their right sides. This verifies the ordered solution (5, 3).

17. Substitution 17: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=-3. Collect to get 2(0)−3y=-3, so −3y=-3 and y=1. Then x=0−(1)=-1.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(1)=0; second is 2(-1)−(1)=-3. Both match their right sides. This verifies the ordered solution (-1, 1).

18. Substitution 18: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 9. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=9. Collect to get 2(3)−3y=9, so −3y=3 and y=-1. Then x=3−(-1)=4.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(-1)=3; second is 2(4)−(-1)=9. Both match their right sides. This verifies the ordered solution (4, -1).

19. Elimination 1: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and x − y = -1. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=-8, so x=-4. Substitute into the first equation: 1(-4)+1y=-7, meaning 1y=-3. Hence y=-3.

Verify in both original equations. Equation one: -4+(-3)=-7. Equation two: -4−(-3)=-1. Both checks hold even when a found value is negative.

20. Elimination 2: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = 10 and 2x − 2y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=12, so x=3. Substitute into the first equation: 2(3)+2y=10, meaning 2y=4. Hence y=2.

Verify in both original equations. Equation one: 6+(4)=10. Equation two: 6−(4)=2. Both checks hold even when a found value is negative.

21. Elimination 3: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = -9 and 3x − 3y = 3. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=-6, so x=-1. Substitute into the first equation: 3(-1)+3y=-9, meaning 3y=-6. Hence y=-2.

Verify in both original equations. Equation one: -3+(-6)=-9. Equation two: -3−(-6)=3. Both checks hold even when a found value is negative.

22. Elimination 4: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 18 and x − 4y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=12, so x=6. Substitute into the first equation: 1(6)+4y=18, meaning 4y=12. Hence y=3.

Verify in both original equations. Equation one: 6+(12)=18. Equation two: 6−(12)=-6. Both checks hold even when a found value is negative.

23. Elimination 5: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 3 and 2x − y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=8, so x=2. Substitute into the first equation: 2(2)+1y=3, meaning 1y=-1. Hence y=-1.

Verify in both original equations. Equation one: 4+(-1)=3. Equation two: 4−(-1)=5. Both checks hold even when a found value is negative.

24. Elimination 6: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 2 and 3x − 2y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=-12, so x=-2. Substitute into the first equation: 3(-2)+2y=2, meaning 2y=8. Hence y=4.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

25. Elimination 7: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = 5 and x − 3y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=10, so x=5. Substitute into the first equation: 1(5)+3y=5, meaning 3y=0. Hence y=0.

Verify in both original equations. Equation one: 5+(0)=5. Equation two: 5−(0)=5. Both checks hold even when a found value is negative.

26. Elimination 8: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 22 and 2x − 4y = -18. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=4, so x=1. Substitute into the first equation: 2(1)+4y=22, meaning 4y=20. Hence y=5.

Verify in both original equations. Equation one: 2+(20)=22. Equation two: 2−(20)=-18. Both checks hold even when a found value is negative.

27. Elimination 9: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + y = -8 and 3x − y = -10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 6x=-18, so x=-3. Substitute into the first equation: 3(-3)+1y=-8, meaning 1y=1. Hence y=1.

Verify in both original equations. Equation one: -9+(1)=-8. Equation two: -9−(1)=-10. Both checks hold even when a found value is negative.

28. Elimination 10: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 2y = -2 and x − 2y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 2x=8, so x=4. Substitute into the first equation: 1(4)+2y=-2, meaning 2y=-6. Hence y=-3.

Verify in both original equations. Equation one: 4+(-6)=-2. Equation two: 4−(-6)=10. Both checks hold even when a found value is negative.

29. Elimination 11: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 3y = 6 and 2x − 3y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 4x=0, so x=0. Substitute into the first equation: 2(0)+3y=6, meaning 3y=6. Hence y=2.

Verify in both original equations. Equation one: 0+(6)=6. Equation two: 0−(6)=-6. Both checks hold even when a found value is negative.

30. Elimination 12: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 4y = -20 and 3x − 4y = -4. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 6x=-24, so x=-4. Substitute into the first equation: 3(-4)+4y=-20, meaning 4y=-8. Hence y=-2.

Verify in both original equations. Equation one: -12+(-8)=-20. Equation two: -12−(-8)=-4. Both checks hold even when a found value is negative.

31. Elimination 13: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and x − y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=6, so x=3. Substitute into the first equation: 1(3)+1y=6, meaning 1y=3. Hence y=3.

Verify in both original equations. Equation one: 3+(3)=6. Equation two: 3−(3)=0. Both checks hold even when a found value is negative.

32. Elimination 14: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = -4 and 2x − 2y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=-4, so x=-1. Substitute into the first equation: 2(-1)+2y=-4, meaning 2y=-2. Hence y=-1.

Verify in both original equations. Equation one: -2+(-2)=-4. Equation two: -2−(-2)=0. Both checks hold even when a found value is negative.

33. Elimination 15: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = 30 and 3x − 3y = 6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=36, so x=6. Substitute into the first equation: 3(6)+3y=30, meaning 3y=12. Hence y=4.

Verify in both original equations. Equation one: 18+(12)=30. Equation two: 18−(12)=6. Both checks hold even when a found value is negative.

34. Elimination 16: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 2 and x − 4y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=4, so x=2. Substitute into the first equation: 1(2)+4y=2, meaning 4y=0. Hence y=0.

Verify in both original equations. Equation one: 2+(0)=2. Equation two: 2−(0)=2. Both checks hold even when a found value is negative.

35. Elimination 17: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 1 and 2x − y = -9. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=-8, so x=-2. Substitute into the first equation: 2(-2)+1y=1, meaning 1y=5. Hence y=5.

Verify in both original equations. Equation one: -4+(5)=1. Equation two: -4−(5)=-9. Both checks hold even when a found value is negative.

36. Elimination 18: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 17 and 3x − 2y = 13. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=30, so x=5. Substitute into the first equation: 3(5)+2y=17, meaning 2y=2. Hence y=1.

Verify in both original equations. Equation one: 15+(2)=17. Equation two: 15−(2)=13. Both checks hold even when a found value is negative.

37. Elimination 19: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = -8 and x − 3y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=2, so x=1. Substitute into the first equation: 1(1)+3y=-8, meaning 3y=-9. Hence y=-3.

Verify in both original equations. Equation one: 1+(-9)=-8. Equation two: 1−(-9)=10. Both checks hold even when a found value is negative.

38. Elimination 20: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 2 and 2x − 4y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=-12, so x=-3. Substitute into the first equation: 2(-3)+4y=2, meaning 4y=8. Hence y=2.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

39. Elimination with multipliers 1: align x-coefficients

Verify in both original equations. The original first equation gives -4+(-5)=-9; the second gives -8+(-5)=-13. Both must be satisfied; a correct value in only one equation is not sufficient.

40. Elimination with multipliers 2: align x-coefficients

Verify in both original equations. The original first equation gives -2+(0)=-2; the second gives -3+(0)=-3. Both must be satisfied; a correct value in only one equation is not sufficient.

41. Elimination with multipliers 3: align x-coefficients

Verify in both original equations. The original first equation gives 6+(25)=31; the second gives 8+(35)=43. Both must be satisfied; a correct value in only one equation is not sufficient.

42. Elimination with multipliers 4: align x-coefficients

Verify in both original equations. The original first equation gives 5+(-2)=3; the second gives 25+(-3)=22. Both must be satisfied; a correct value in only one equation is not sufficient.

43. Elimination with multipliers 5: align x-coefficients

Verify in both original equations. The original first equation gives -6+(16)=10; the second gives -6+(24)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

44. Elimination with multipliers 6: align x-coefficients

Verify in both original equations. The original first equation gives 0+(-2)=-2; the second gives 0+(-4)=-4. Both must be satisfied; a correct value in only one equation is not sufficient.

45. Elimination with multipliers 7: align x-coefficients

Verify in both original equations. The original first equation gives 3+(9)=12; the second gives 12+(15)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

46. Elimination with multipliers 8: align x-coefficients

Verify in both original equations. The original first equation gives 12+(-15)=-3; the second gives 30+(-3)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

47. Elimination with multipliers 9: align x-coefficients

Verify in both original equations. The original first equation gives -6+(4)=-2; the second gives -4+(8)=4. Both must be satisfied; a correct value in only one equation is not sufficient.

48. Elimination with multipliers 10: align x-coefficients

Verify in both original equations. The original first equation gives 1+(-16)=-15; the second gives 3+(-28)=-25. Both must be satisfied; a correct value in only one equation is not sufficient.

49. Elimination with multipliers 11: align x-coefficients

Verify in both original equations. The original first equation gives 8+(1)=9; the second gives 16+(3)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

50. Elimination with multipliers 12: align x-coefficients

Verify in both original equations. The original first equation gives -12+(-15)=-27; the second gives -20+(-30)=-50. Both must be satisfied; a correct value in only one equation is not sufficient.

51. Elimination with multipliers 13: align x-coefficients

Verify in both original equations. The original first equation gives -1+(0)=-1; the second gives -2+(0)=-2. Both must be satisfied; a correct value in only one equation is not sufficient.

52. Elimination with multipliers 14: align x-coefficients

Verify in both original equations. The original first equation gives 4+(10)=14; the second gives 6+(25)=31. Both must be satisfied; a correct value in only one equation is not sufficient.

53. Elimination with multipliers 15: align x-coefficients

Verify in both original equations. The original first equation gives 15+(-4)=11; the second gives 20+(-1)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

54. Elimination with multipliers 16: align x-coefficients

Verify in both original equations. The original first equation gives -3+(4)=1; the second gives -15+(16)=1. Both must be satisfied; a correct value in only one equation is not sufficient.

55. Elimination with multipliers 17: align x-coefficients

Verify in both original equations. The original first equation gives 0+(-6)=-6; the second gives 0+(-14)=-14. Both must be satisfied; a correct value in only one equation is not sufficient.

56. Elimination with multipliers 18: align x-coefficients

Verify in both original equations. The original first equation gives 9+(15)=24; the second gives 9+(9)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

57. Elimination with multipliers 19: align x-coefficients

Verify in both original equations. The original first equation gives 6+(-6)=0; the second gives 24+(-18)=6. Both must be satisfied; a correct value in only one equation is not sufficient.

58. Elimination with multipliers 20: align x-coefficients

Verify in both original equations. The original first equation gives -4+(8)=4; the second gives -10+(4)=-6. Both must be satisfied; a correct value in only one equation is not sufficient.

59. Elimination with multipliers 21: align x-coefficients

Verify in both original equations. The original first equation gives 3+(-4)=-1; the second gives 2+(-20)=-18. Both must be satisfied; a correct value in only one equation is not sufficient.

60. Elimination with multipliers 22: align x-coefficients

Verify in both original equations. The original first equation gives 4+(3)=7; the second gives 12+(1)=13. Both must be satisfied; a correct value in only one equation is not sufficient.

61. Elimination with multipliers 23: align x-coefficients

Verify in both original equations. The original first equation gives -8+(-25)=-33; the second gives -16+(-20)=-36. Both must be satisfied; a correct value in only one equation is not sufficient.

62. Elimination with multipliers 24: align x-coefficients

Verify in both original equations. The original first equation gives -3+(0)=-3; the second gives -5+(0)=-5. Both must be satisfied; a correct value in only one equation is not sufficient.

63. Fictional purchase model 1: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×2+1×2=$6, and the second 1×2+3×2=$8. The simultaneous system reconstructs both supplied totals.

64. Fictional purchase model 2: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×3+2×5=$13, and the second 3×3+1×5=$14. The simultaneous system reconstructs both supplied totals.

65. Fictional purchase model 3: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×4+2×8=$28, and the second 1×4+4×8=$36. The simultaneous system reconstructs both supplied totals.

66. Fictional purchase model 4: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×5+3×4=$22, and the second 4×5+1×4=$24. The simultaneous system reconstructs both supplied totals.

67. Fictional purchase model 5: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×6+3×7=$27, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

68. Fictional purchase model 6: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×7+4×3=$26, and the second 1×7+3×3=$16. The simultaneous system reconstructs both supplied totals.

69. Fictional purchase model 7: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×8+1×6=$30, and the second 2×8+5×6=$46. The simultaneous system reconstructs both supplied totals.

70. Fictional purchase model 8: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×2+4×2=$10, and the second 3×2+2×2=$10. The simultaneous system reconstructs both supplied totals.

71. Fictional purchase model 9: two unknown prices

Verify in both original equations. At these prices, the first basket costs 4×3+1×5=$17, and the second 2×3+3×5=$21. The simultaneous system reconstructs both supplied totals.

72. Fictional purchase model 10: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×4+5×8=$48, and the second 3×4+1×8=$20. The simultaneous system reconstructs both supplied totals.

73. Fictional purchase model 11: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×5+4×4=$31, and the second 1×5+2×4=$13. The simultaneous system reconstructs both supplied totals.

74. Fictional purchase model 12: two unknown prices

Verify in both original equations. At these prices, the first basket costs 4×6+3×7=$45, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

75. Graphical interpretation 1: intersection of two lines

Worked method. Set the right sides equal: 1x+(-1)=2x+(2). Thus (-1)x=3, giving x=-3. Substitute in either rule to get y=-4. The graphs intersect at (-3, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: 1(-3)+(-1)=-4. Second: 2(-3)+(2)=-4. The lines have distinct slopes 1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

76. Graphical interpretation 2: intersection of two lines

Worked method. Set the right sides equal: 1x+(3)=-1x+(-1). Thus (2)x=-4, giving x=-2. Substitute in either rule to get y=1. The graphs intersect at (-2, 1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 1(-2)+(3)=1. Second: -1(-2)+(-1)=1. The lines have distinct slopes 1 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

77. Graphical interpretation 3: intersection of two lines

Worked method. Set the right sides equal: 2x+(8)=-2x+(4). Thus (4)x=-4, giving x=-1. Substitute in either rule to get y=6. The graphs intersect at (-1, 6).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 2(-1)+(8)=6. Second: -2(-1)+(4)=6. The lines have distinct slopes 2 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

78. Graphical interpretation 4: intersection of two lines

Worked method. Set the right sides equal: 3x+(0)=1x+(0). Thus (2)x=0, giving x=0. Substitute in either rule to get y=0. The graphs intersect at (0, 0).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=0: 3(0)+(0)=0. Second: 1(0)+(0)=0. The lines have distinct slopes 3 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

79. Graphical interpretation 5: intersection of two lines

Worked method. Set the right sides equal: -1x+(6)=2x+(3). Thus (-3)x=-3, giving x=1. Substitute in either rule to get y=5. The graphs intersect at (1, 5).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=1: -1(1)+(6)=5. Second: 2(1)+(3)=5. The lines have distinct slopes -1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

80. Graphical interpretation 6: intersection of two lines

Worked method. Set the right sides equal: -2x+(3)=1x+(-3). Thus (-3)x=-6, giving x=2. Substitute in either rule to get y=-1. The graphs intersect at (2, -1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=2: -2(2)+(3)=-1. Second: 1(2)+(-3)=-1. The lines have distinct slopes -2 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

81. Graphical interpretation 7: intersection of two lines

Worked method. Set the right sides equal: 2x+(-2)=3x+(-5). Thus (-1)x=-3, giving x=3. Substitute in either rule to get y=4. The graphs intersect at (3, 4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=3: 2(3)+(-2)=4. Second: 3(3)+(-5)=4. The lines have distinct slopes 2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

82. Graphical interpretation 8: intersection of two lines

Worked method. Set the right sides equal: -1x+(2)=-3x+(10). Thus (2)x=8, giving x=4. Substitute in either rule to get y=-2. The graphs intersect at (4, -2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=4: -1(4)+(2)=-2. Second: -3(4)+(10)=-2. The lines have distinct slopes -1 and -3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

83. Graphical interpretation 9: intersection of two lines

Worked method. Set the right sides equal: 1x+(-2)=4x+(-17). Thus (-3)x=-15, giving x=5. Substitute in either rule to get y=3. The graphs intersect at (5, 3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=5: 1(5)+(-2)=3. Second: 4(5)+(-17)=3. The lines have distinct slopes 1 and 4, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

84. Graphical interpretation 10: intersection of two lines

Worked method. Set the right sides equal: -2x+(-9)=3x+(6). Thus (-5)x=15, giving x=-3. Substitute in either rule to get y=-3. The graphs intersect at (-3, -3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: -2(-3)+(-9)=-3. Second: 3(-3)+(6)=-3. The lines have distinct slopes -2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

85. Graphical interpretation 11: intersection of two lines

Worked method. Set the right sides equal: 3x+(8)=-2x+(-2). Thus (5)x=-10, giving x=-2. Substitute in either rule to get y=2. The graphs intersect at (-2, 2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 3(-2)+(8)=2. Second: -2(-2)+(-2)=2. The lines have distinct slopes 3 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

86. Graphical interpretation 12: intersection of two lines

Worked method. Set the right sides equal: 4x+(0)=-1x+(-5). Thus (5)x=-5, giving x=-1. Substitute in either rule to get y=-4. The graphs intersect at (-1, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 4(-1)+(0)=-4. Second: -1(-1)+(-5)=-4. The lines have distinct slopes 4 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

Five different errors that may look like one wrong answer

Only one equation is satisfied

Brackets are lost during substitution

Only one side is multiplied in elimination

A subtraction sign is reversed inconsistently

A graph’s y-intercept is confused with intersection

A three-pupil tutorial can support different readiness levels

10 minutes: retrieve one-variable equation balance and ordered-pair plotting.
20 minutes: tutor models substitution from a straightforward linear condition.
25 minutes: solve differentiated systems and diagnose each student’s first wrong line.
15 minutes: compare elimination with graphical interpretation.
10 minutes: independently solve an unseen changed-number system.
10 minutes: check both original equations and assign a delayed retest.

A practical five-session bridge for parents

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Did you know? One equation in two unknowns often gives many possible answers. A second independent equation narrows the possibilities to the pair that satisfies both. That is the core idea behind simultaneous linear equations: two conditions, one shared solution. Students who already understand Secondary 2 G1 one-variable equation balance can use this as a carefully staged bridge toward the full G1 Mathematics syllabus.

Quick answer: three ways to solve simultaneous equations

Substitution replaces one variable using a rearranged equation. Elimination adds or subtracts equations (sometimes after multiplying every term) until one variable disappears. The graphical method treats each equation as a line and finds their common intersection point. Whatever method is chosen, substitute both found values back into both original equations to verify the pair.

Syllabus-year distinction: The MOE 2020 Secondary Two G1 syllabus includes solving and forming linear equations in one variable. The full 2027 SEC K110 G1 Mathematics syllabus includes simultaneous linear equations in two variables by substitution, elimination and the graphical method. The worked cases in this article are a later-stage bridge for a ready learner, not an assertion that these methods must appear in a 2026 Secondary 2 G1 school assessment.

Choose the relevant explanation

A one-minute prerequisite diagnostic
Substitution: use one equation inside another
Elimination: remove one unknown
Graphical method and intersections
More than 80 original worked clinics
Five common first-line errors
Small-group tutorial model
A parent-friendly practice schedule
FAQs
Official sources

A 60-second check before teaching two unknowns

Substitution: make one variable speak through the other

Elimination: preserve equality while removing one letter

Graphical solutions connect back to Secondary 2 coordinates

86 original simultaneous-equation learning clinics

1. Substitution 1: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(-3)=-7; second is 2(-4)−(-3)=-5. Both match their right sides. This verifies the ordered solution (-4, -3).

2. Substitution 2: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-2. Collect to get 2(5)−3y=-2, so −3y=-12 and y=4. Then x=5−(4)=1.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(4)=5; second is 2(1)−(4)=-2. Both match their right sides. This verifies the ordered solution (1, 4).

3. Substitution 3: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 10. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=10. Collect to get 2(8)−3y=10, so −3y=-6 and y=2. Then x=8−(2)=6.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(2)=8; second is 2(6)−(2)=10. Both match their right sides. This verifies the ordered solution (6, 2).

4. Substitution 4: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=0. Collect to get 2(0)−3y=0, so −3y=0 and y=0. Then x=0−(0)=0.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(0)=0; second is 2(0)−(0)=0. Both match their right sides. This verifies the ordered solution (0, 0).

5. Substitution 5: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 12. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=12. Collect to get 2(3)−3y=12, so −3y=6 and y=-2. Then x=3−(-2)=5.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(-2)=3; second is 2(5)−(-2)=12. Both match their right sides. This verifies the ordered solution (5, -2).

6. Substitution 6: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = -7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=-7. Collect to get 2(4)−3y=-7, so −3y=-15 and y=5. Then x=4−(5)=-1.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(5)=4; second is 2(-1)−(5)=-7. Both match their right sides. This verifies the ordered solution (-1, 5).

7. Substitution 7: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 7 and 2x − y = 5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=7−y. Substitute this entire expression for x in the second: 2(7−y)−y=5. Collect to get 2(7)−3y=5, so −3y=-9 and y=3. Then x=7−(3)=4.

First wrong step to avoid. A learner may replace x with 7 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(3)=7; second is 2(4)−(3)=5. Both match their right sides. This verifies the ordered solution (4, 3).

8. Substitution 8: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -1 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -2+(1)=-1; second is 2(-2)−(1)=-5. Both match their right sides. This verifies the ordered solution (-2, 1).

9. Substitution 9: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 2 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=2−y. Substitute this entire expression for x in the second: 2(2−y)−y=7. Collect to get 2(2)−3y=7, so −3y=3 and y=-1. Then x=2−(-1)=3.

First wrong step to avoid. A learner may replace x with 2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 3+(-1)=2; second is 2(3)−(-1)=7. Both match their right sides. This verifies the ordered solution (3, -1).

10. Substitution 10: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -6 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -3+(-3)=-6; second is 2(-3)−(-3)=-3. Both match their right sides. This verifies the ordered solution (-3, -3).

11. Substitution 11: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and 2x − y = 0. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=6−y. Substitute this entire expression for x in the second: 2(6−y)−y=0. Collect to get 2(6)−3y=0, so −3y=-12 and y=4. Then x=6−(4)=2.

First wrong step to avoid. A learner may replace x with 6 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 2+(4)=6; second is 2(2)−(4)=0. Both match their right sides. This verifies the ordered solution (2, 4).

12. Substitution 12: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -2 and 2x − y = -10. Find both unknowns, not merely the value of x+y.

First wrong step to avoid. A learner may replace x with -2 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -4+(2)=-2; second is 2(-4)−(2)=-10. Both match their right sides. This verifies the ordered solution (-4, 2).

13. Substitution 13: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 1 and 2x − y = 2. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=1−y. Substitute this entire expression for x in the second: 2(1−y)−y=2. Collect to get 2(1)−3y=2, so −3y=0 and y=0. Then x=1−(0)=1.

First wrong step to avoid. A learner may replace x with 1 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 1+(0)=1; second is 2(1)−(0)=2. Both match their right sides. This verifies the ordered solution (1, 0).

14. Substitution 14: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 4 and 2x − y = 14. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=4−y. Substitute this entire expression for x in the second: 2(4−y)−y=14. Collect to get 2(4)−3y=14, so −3y=6 and y=-2. Then x=4−(-2)=6.

First wrong step to avoid. A learner may replace x with 4 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 6+(-2)=4; second is 2(6)−(-2)=14. Both match their right sides. This verifies the ordered solution (6, -2).

15. Substitution 15: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 5 and 2x − y = -5. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=5−y. Substitute this entire expression for x in the second: 2(5−y)−y=-5. Collect to get 2(5)−3y=-5, so −3y=-15 and y=5. Then x=5−(5)=0.

First wrong step to avoid. A learner may replace x with 5 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 0+(5)=5; second is 2(0)−(5)=-5. Both match their right sides. This verifies the ordered solution (0, 5).

16. Substitution 16: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 8 and 2x − y = 7. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=8−y. Substitute this entire expression for x in the second: 2(8−y)−y=7. Collect to get 2(8)−3y=7, so −3y=-9 and y=3. Then x=8−(3)=5.

First wrong step to avoid. A learner may replace x with 8 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 5+(3)=8; second is 2(5)−(3)=7. Both match their right sides. This verifies the ordered solution (5, 3).

17. Substitution 17: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 0 and 2x − y = -3. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=0−y. Substitute this entire expression for x in the second: 2(0−y)−y=-3. Collect to get 2(0)−3y=-3, so −3y=-3 and y=1. Then x=0−(1)=-1.

First wrong step to avoid. A learner may replace x with 0 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side -1+(1)=0; second is 2(-1)−(1)=-3. Both match their right sides. This verifies the ordered solution (-1, 1).

18. Substitution 18: isolate a variable

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 3 and 2x − y = 9. Find both unknowns, not merely the value of x+y.

Worked method. From the first equation, x=3−y. Substitute this entire expression for x in the second: 2(3−y)−y=9. Collect to get 2(3)−3y=9, so −3y=3 and y=-1. Then x=3−(-1)=4.

First wrong step to avoid. A learner may replace x with 3 but forget its remaining −y, or may stop after finding one coordinate. Both equations must be satisfied by the same pair.

Verify in both original equations. First left side 4+(-1)=3; second is 2(4)−(-1)=9. Both match their right sides. This verifies the ordered solution (4, -1).

19. Elimination 1: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = -7 and x − y = -1. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=-8, so x=-4. Substitute into the first equation: 1(-4)+1y=-7, meaning 1y=-3. Hence y=-3.

Verify in both original equations. Equation one: -4+(-3)=-7. Equation two: -4−(-3)=-1. Both checks hold even when a found value is negative.

20. Elimination 2: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = 10 and 2x − 2y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=12, so x=3. Substitute into the first equation: 2(3)+2y=10, meaning 2y=4. Hence y=2.

Verify in both original equations. Equation one: 6+(4)=10. Equation two: 6−(4)=2. Both checks hold even when a found value is negative.

21. Elimination 3: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = -9 and 3x − 3y = 3. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=-6, so x=-1. Substitute into the first equation: 3(-1)+3y=-9, meaning 3y=-6. Hence y=-2.

Verify in both original equations. Equation one: -3+(-6)=-9. Equation two: -3−(-6)=3. Both checks hold even when a found value is negative.

22. Elimination 4: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 18 and x − 4y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=12, so x=6. Substitute into the first equation: 1(6)+4y=18, meaning 4y=12. Hence y=3.

Verify in both original equations. Equation one: 6+(12)=18. Equation two: 6−(12)=-6. Both checks hold even when a found value is negative.

23. Elimination 5: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 3 and 2x − y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=8, so x=2. Substitute into the first equation: 2(2)+1y=3, meaning 1y=-1. Hence y=-1.

Verify in both original equations. Equation one: 4+(-1)=3. Equation two: 4−(-1)=5. Both checks hold even when a found value is negative.

24. Elimination 6: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 2 and 3x − 2y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=-12, so x=-2. Substitute into the first equation: 3(-2)+2y=2, meaning 2y=8. Hence y=4.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

25. Elimination 7: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = 5 and x − 3y = 5. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=10, so x=5. Substitute into the first equation: 1(5)+3y=5, meaning 3y=0. Hence y=0.

Verify in both original equations. Equation one: 5+(0)=5. Equation two: 5−(0)=5. Both checks hold even when a found value is negative.

26. Elimination 8: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 22 and 2x − 4y = -18. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=4, so x=1. Substitute into the first equation: 2(1)+4y=22, meaning 4y=20. Hence y=5.

Verify in both original equations. Equation one: 2+(20)=22. Equation two: 2−(20)=-18. Both checks hold even when a found value is negative.

27. Elimination 9: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + y = -8 and 3x − y = -10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 6x=-18, so x=-3. Substitute into the first equation: 3(-3)+1y=-8, meaning 1y=1. Hence y=1.

Verify in both original equations. Equation one: -9+(1)=-8. Equation two: -9−(1)=-10. Both checks hold even when a found value is negative.

28. Elimination 10: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 2y = -2 and x − 2y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 2x=8, so x=4. Substitute into the first equation: 1(4)+2y=-2, meaning 2y=-6. Hence y=-3.

Verify in both original equations. Equation one: 4+(-6)=-2. Equation two: 4−(-6)=10. Both checks hold even when a found value is negative.

29. Elimination 11: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 3y = 6 and 2x − 3y = -6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 4x=0, so x=0. Substitute into the first equation: 2(0)+3y=6, meaning 3y=6. Hence y=2.

Verify in both original equations. Equation one: 0+(6)=6. Equation two: 0−(6)=-6. Both checks hold even when a found value is negative.

30. Elimination 12: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 4y = -20 and 3x − 4y = -4. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 6x=-24, so x=-4. Substitute into the first equation: 3(-4)+4y=-20, meaning 4y=-8. Hence y=-2.

Verify in both original equations. Equation one: -12+(-8)=-20. Equation two: -12−(-8)=-4. Both checks hold even when a found value is negative.

31. Elimination 13: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + y = 6 and x − y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 2x=6, so x=3. Substitute into the first equation: 1(3)+1y=6, meaning 1y=3. Hence y=3.

Verify in both original equations. Equation one: 3+(3)=6. Equation two: 3−(3)=0. Both checks hold even when a found value is negative.

32. Elimination 14: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 2y = -4 and 2x − 2y = 0. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 4x=-4, so x=-1. Substitute into the first equation: 2(-1)+2y=-4, meaning 2y=-2. Hence y=-1.

Verify in both original equations. Equation one: -2+(-2)=-4. Equation two: -2−(-2)=0. Both checks hold even when a found value is negative.

33. Elimination 15: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 3y = 30 and 3x − 3y = 6. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 6x=36, so x=6. Substitute into the first equation: 3(6)+3y=30, meaning 3y=12. Hence y=4.

Verify in both original equations. Equation one: 18+(12)=30. Equation two: 18−(12)=6. Both checks hold even when a found value is negative.

34. Elimination 16: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 4y = 2 and x − 4y = 2. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 2x=4, so x=2. Substitute into the first equation: 1(2)+4y=2, meaning 4y=0. Hence y=0.

Verify in both original equations. Equation one: 2+(0)=2. Equation two: 2−(0)=2. Both checks hold even when a found value is negative.

35. Elimination 17: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + y = 1 and 2x − y = -9. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +1y and −1y terms cancel. This leaves 4x=-8, so x=-2. Substitute into the first equation: 2(-2)+1y=1, meaning 1y=5. Hence y=5.

Verify in both original equations. Equation one: -4+(5)=1. Equation two: -4−(5)=-9. Both checks hold even when a found value is negative.

36. Elimination 18: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 3x + 2y = 17 and 3x − 2y = 13. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +2y and −2y terms cancel. This leaves 6x=30, so x=5. Substitute into the first equation: 3(5)+2y=17, meaning 2y=2. Hence y=1.

Verify in both original equations. Equation one: 15+(2)=17. Equation two: 15−(2)=13. Both checks hold even when a found value is negative.

37. Elimination 19: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve x + 3y = -8 and x − 3y = 10. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +3y and −3y terms cancel. This leaves 2x=2, so x=1. Substitute into the first equation: 1(1)+3y=-8, meaning 3y=-9. Hence y=-3.

Verify in both original equations. Equation one: 1+(-9)=-8. Equation two: 1−(-9)=10. Both checks hold even when a found value is negative.

38. Elimination 20: cancel opposite y-terms

Did you know? A simultaneous solution must satisfy two conditions at once. Solve 2x + 4y = 2 and 2x − 4y = -14. The y terms have equal magnitude and opposite signs.

Worked method. Add the equations: the +4y and −4y terms cancel. This leaves 4x=-12, so x=-3. Substitute into the first equation: 2(-3)+4y=2, meaning 4y=8. Hence y=2.

Verify in both original equations. Equation one: -6+(8)=2. Equation two: -6−(8)=-14. Both checks hold even when a found value is negative.

39. Elimination with multipliers 1: align x-coefficients

Verify in both original equations. The original first equation gives -4+(-5)=-9; the second gives -8+(-5)=-13. Both must be satisfied; a correct value in only one equation is not sufficient.

40. Elimination with multipliers 2: align x-coefficients

Verify in both original equations. The original first equation gives -2+(0)=-2; the second gives -3+(0)=-3. Both must be satisfied; a correct value in only one equation is not sufficient.

41. Elimination with multipliers 3: align x-coefficients

Verify in both original equations. The original first equation gives 6+(25)=31; the second gives 8+(35)=43. Both must be satisfied; a correct value in only one equation is not sufficient.

42. Elimination with multipliers 4: align x-coefficients

Verify in both original equations. The original first equation gives 5+(-2)=3; the second gives 25+(-3)=22. Both must be satisfied; a correct value in only one equation is not sufficient.

43. Elimination with multipliers 5: align x-coefficients

Verify in both original equations. The original first equation gives -6+(16)=10; the second gives -6+(24)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

44. Elimination with multipliers 6: align x-coefficients

Verify in both original equations. The original first equation gives 0+(-2)=-2; the second gives 0+(-4)=-4. Both must be satisfied; a correct value in only one equation is not sufficient.

45. Elimination with multipliers 7: align x-coefficients

Verify in both original equations. The original first equation gives 3+(9)=12; the second gives 12+(15)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

46. Elimination with multipliers 8: align x-coefficients

Verify in both original equations. The original first equation gives 12+(-15)=-3; the second gives 30+(-3)=27. Both must be satisfied; a correct value in only one equation is not sufficient.

47. Elimination with multipliers 9: align x-coefficients

Verify in both original equations. The original first equation gives -6+(4)=-2; the second gives -4+(8)=4. Both must be satisfied; a correct value in only one equation is not sufficient.

48. Elimination with multipliers 10: align x-coefficients

Verify in both original equations. The original first equation gives 1+(-16)=-15; the second gives 3+(-28)=-25. Both must be satisfied; a correct value in only one equation is not sufficient.

49. Elimination with multipliers 11: align x-coefficients

Verify in both original equations. The original first equation gives 8+(1)=9; the second gives 16+(3)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

50. Elimination with multipliers 12: align x-coefficients

Verify in both original equations. The original first equation gives -12+(-15)=-27; the second gives -20+(-30)=-50. Both must be satisfied; a correct value in only one equation is not sufficient.

51. Elimination with multipliers 13: align x-coefficients

Verify in both original equations. The original first equation gives -1+(0)=-1; the second gives -2+(0)=-2. Both must be satisfied; a correct value in only one equation is not sufficient.

52. Elimination with multipliers 14: align x-coefficients

Verify in both original equations. The original first equation gives 4+(10)=14; the second gives 6+(25)=31. Both must be satisfied; a correct value in only one equation is not sufficient.

53. Elimination with multipliers 15: align x-coefficients

Verify in both original equations. The original first equation gives 15+(-4)=11; the second gives 20+(-1)=19. Both must be satisfied; a correct value in only one equation is not sufficient.

54. Elimination with multipliers 16: align x-coefficients

Verify in both original equations. The original first equation gives -3+(4)=1; the second gives -15+(16)=1. Both must be satisfied; a correct value in only one equation is not sufficient.

55. Elimination with multipliers 17: align x-coefficients

Verify in both original equations. The original first equation gives 0+(-6)=-6; the second gives 0+(-14)=-14. Both must be satisfied; a correct value in only one equation is not sufficient.

56. Elimination with multipliers 18: align x-coefficients

Verify in both original equations. The original first equation gives 9+(15)=24; the second gives 9+(9)=18. Both must be satisfied; a correct value in only one equation is not sufficient.

57. Elimination with multipliers 19: align x-coefficients

Verify in both original equations. The original first equation gives 6+(-6)=0; the second gives 24+(-18)=6. Both must be satisfied; a correct value in only one equation is not sufficient.

58. Elimination with multipliers 20: align x-coefficients

Verify in both original equations. The original first equation gives -4+(8)=4; the second gives -10+(4)=-6. Both must be satisfied; a correct value in only one equation is not sufficient.

59. Elimination with multipliers 21: align x-coefficients

Verify in both original equations. The original first equation gives 3+(-4)=-1; the second gives 2+(-20)=-18. Both must be satisfied; a correct value in only one equation is not sufficient.

60. Elimination with multipliers 22: align x-coefficients

Verify in both original equations. The original first equation gives 4+(3)=7; the second gives 12+(1)=13. Both must be satisfied; a correct value in only one equation is not sufficient.

61. Elimination with multipliers 23: align x-coefficients

Verify in both original equations. The original first equation gives -8+(-25)=-33; the second gives -16+(-20)=-36. Both must be satisfied; a correct value in only one equation is not sufficient.

62. Elimination with multipliers 24: align x-coefficients

Verify in both original equations. The original first equation gives -3+(0)=-3; the second gives -5+(0)=-5. Both must be satisfied; a correct value in only one equation is not sufficient.

63. Fictional purchase model 1: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×2+1×2=$6, and the second 1×2+3×2=$8. The simultaneous system reconstructs both supplied totals.

64. Fictional purchase model 2: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×3+2×5=$13, and the second 3×3+1×5=$14. The simultaneous system reconstructs both supplied totals.

65. Fictional purchase model 3: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×4+2×8=$28, and the second 1×4+4×8=$36. The simultaneous system reconstructs both supplied totals.

66. Fictional purchase model 4: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×5+3×4=$22, and the second 4×5+1×4=$24. The simultaneous system reconstructs both supplied totals.

67. Fictional purchase model 5: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×6+3×7=$27, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

68. Fictional purchase model 6: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×7+4×3=$26, and the second 1×7+3×3=$16. The simultaneous system reconstructs both supplied totals.

69. Fictional purchase model 7: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×8+1×6=$30, and the second 2×8+5×6=$46. The simultaneous system reconstructs both supplied totals.

70. Fictional purchase model 8: two unknown prices

Verify in both original equations. At these prices, the first basket costs 1×2+4×2=$10, and the second 3×2+2×2=$10. The simultaneous system reconstructs both supplied totals.

71. Fictional purchase model 9: two unknown prices

Verify in both original equations. At these prices, the first basket costs 4×3+1×5=$17, and the second 2×3+3×5=$21. The simultaneous system reconstructs both supplied totals.

72. Fictional purchase model 10: two unknown prices

Verify in both original equations. At these prices, the first basket costs 2×4+5×8=$48, and the second 3×4+1×8=$20. The simultaneous system reconstructs both supplied totals.

73. Fictional purchase model 11: two unknown prices

Verify in both original equations. At these prices, the first basket costs 3×5+4×4=$31, and the second 1×5+2×4=$13. The simultaneous system reconstructs both supplied totals.

74. Fictional purchase model 12: two unknown prices

Verify in both original equations. At these prices, the first basket costs 4×6+3×7=$45, and the second 2×6+1×7=$19. The simultaneous system reconstructs both supplied totals.

75. Graphical interpretation 1: intersection of two lines

Worked method. Set the right sides equal: 1x+(-1)=2x+(2). Thus (-1)x=3, giving x=-3. Substitute in either rule to get y=-4. The graphs intersect at (-3, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: 1(-3)+(-1)=-4. Second: 2(-3)+(2)=-4. The lines have distinct slopes 1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

76. Graphical interpretation 2: intersection of two lines

Worked method. Set the right sides equal: 1x+(3)=-1x+(-1). Thus (2)x=-4, giving x=-2. Substitute in either rule to get y=1. The graphs intersect at (-2, 1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 1(-2)+(3)=1. Second: -1(-2)+(-1)=1. The lines have distinct slopes 1 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

77. Graphical interpretation 3: intersection of two lines

Worked method. Set the right sides equal: 2x+(8)=-2x+(4). Thus (4)x=-4, giving x=-1. Substitute in either rule to get y=6. The graphs intersect at (-1, 6).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 2(-1)+(8)=6. Second: -2(-1)+(4)=6. The lines have distinct slopes 2 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

78. Graphical interpretation 4: intersection of two lines

Worked method. Set the right sides equal: 3x+(0)=1x+(0). Thus (2)x=0, giving x=0. Substitute in either rule to get y=0. The graphs intersect at (0, 0).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=0: 3(0)+(0)=0. Second: 1(0)+(0)=0. The lines have distinct slopes 3 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

79. Graphical interpretation 5: intersection of two lines

Worked method. Set the right sides equal: -1x+(6)=2x+(3). Thus (-3)x=-3, giving x=1. Substitute in either rule to get y=5. The graphs intersect at (1, 5).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=1: -1(1)+(6)=5. Second: 2(1)+(3)=5. The lines have distinct slopes -1 and 2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

80. Graphical interpretation 6: intersection of two lines

Worked method. Set the right sides equal: -2x+(3)=1x+(-3). Thus (-3)x=-6, giving x=2. Substitute in either rule to get y=-1. The graphs intersect at (2, -1).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=2: -2(2)+(3)=-1. Second: 1(2)+(-3)=-1. The lines have distinct slopes -2 and 1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

81. Graphical interpretation 7: intersection of two lines

Worked method. Set the right sides equal: 2x+(-2)=3x+(-5). Thus (-1)x=-3, giving x=3. Substitute in either rule to get y=4. The graphs intersect at (3, 4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=3: 2(3)+(-2)=4. Second: 3(3)+(-5)=4. The lines have distinct slopes 2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

82. Graphical interpretation 8: intersection of two lines

Worked method. Set the right sides equal: -1x+(2)=-3x+(10). Thus (2)x=8, giving x=4. Substitute in either rule to get y=-2. The graphs intersect at (4, -2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=4: -1(4)+(2)=-2. Second: -3(4)+(10)=-2. The lines have distinct slopes -1 and -3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

83. Graphical interpretation 9: intersection of two lines

Worked method. Set the right sides equal: 1x+(-2)=4x+(-17). Thus (-3)x=-15, giving x=5. Substitute in either rule to get y=3. The graphs intersect at (5, 3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=5: 1(5)+(-2)=3. Second: 4(5)+(-17)=3. The lines have distinct slopes 1 and 4, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

84. Graphical interpretation 10: intersection of two lines

Worked method. Set the right sides equal: -2x+(-9)=3x+(6). Thus (-5)x=15, giving x=-3. Substitute in either rule to get y=-3. The graphs intersect at (-3, -3).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-3: -2(-3)+(-9)=-3. Second: 3(-3)+(6)=-3. The lines have distinct slopes -2 and 3, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

85. Graphical interpretation 11: intersection of two lines

Worked method. Set the right sides equal: 3x+(8)=-2x+(-2). Thus (5)x=-10, giving x=-2. Substitute in either rule to get y=2. The graphs intersect at (-2, 2).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-2: 3(-2)+(8)=2. Second: -2(-2)+(-2)=2. The lines have distinct slopes 3 and -2, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

86. Graphical interpretation 12: intersection of two lines

Worked method. Set the right sides equal: 4x+(0)=-1x+(-5). Thus (5)x=-5, giving x=-1. Substitute in either rule to get y=-4. The graphs intersect at (-1, -4).

First wrong step to avoid. A learner may confuse the point where each line crosses the y-axis with where the two lines cross one another. A simultaneous solution belongs to both graphs at once.

Verify in both original equations. First output at x=-1: 4(-1)+(0)=-4. Second: -1(-1)+(-5)=-4. The lines have distinct slopes 4 and -1, so this intersection is unique.

Transfer task. Sketch the two straight lines with labelled scales. Shift only one intercept and predict how the intersection changes, then verify by solving the new equal-output equation.

Five different errors that may look like one wrong answer

Only one equation is satisfied

Brackets are lost during substitution

Only one side is multiplied in elimination

A subtraction sign is reversed inconsistently

A graph’s y-intercept is confused with intersection

A three-pupil tutorial can support different readiness levels

10 minutes: retrieve one-variable equation balance and ordered-pair plotting.
20 minutes: tutor models substitution from a straightforward linear condition.
25 minutes: solve differentiated systems and diagnose each student’s first wrong line.
15 minutes: compare elimination with graphical interpretation.
10 minutes: independently solve an unseen changed-number system.
10 minutes: check both original equations and assign a delayed retest.

A practical five-session bridge for parents

Session one: review x+y=constant and one-variable linear equations.
Session two: solve two substitution cases using explicit replacement brackets.
Session three: solve two elimination cases and verify every term of the scaled equation.
Session four: sketch two lines and interpret their intersection using labelled axes.
Session five: explain a short two-price story with variables and check both original totals.

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

It is often convenient when one variable can be isolated simply, such as x+y=9 or y=2x+3.

When is elimination more convenient?

When one variable has coefficients that already match, or can be made to match by multiplying complete equations.

Why must an answer be a pair rather than a single number?

Both unknowns need values that satisfy the two conditions together. An ordered pair communicates the joint solution.

Can a system have no solution?

Yes. Distinct parallel lines have no shared point. Two equivalent equations describe the same line and have infinitely many common points.

How can a tutor check mastery without a full test?

Choose one system the student has not seen, ask them to select a method, show valid working and check both original equations independently.

Official K110 content and reader note

For a suitable learning plan, take one marked one-variable equation problem to eduKate Singapore and ask whether the pupil should rebuild equality skills before attempting two-unknown systems.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Simultaneous Linear Equations Explained

Official K110 content and reader note

Official K110 content and reader note

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

When is elimination more convenient?

Why must an answer be a pair rather than a single number?

Can a system have no solution?

How can a tutor check mastery without a full test?

Continue from the Secondary 2 foundation

Official K110 content and reader note

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

When is elimination more convenient?

Why must an answer be a pair rather than a single number?

Can a system have no solution?

How can a tutor check mastery without a full test?

Continue from the Secondary 2 foundation

Official K110 content and reader note

A practical five-session bridge for parents

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

When is elimination more convenient?

Why must an answer be a pair rather than a single number?

Can a system have no solution?

How can a tutor check mastery without a full test?

Continue from the Secondary 2 foundation

Official K110 content and reader note

A practical five-session bridge for parents

Frequently asked questions

Are simultaneous linear equations a mandatory Secondary 2 G1 topic?

When should substitution be chosen?

When is elimination more convenient?

Why must an answer be a pair rather than a single number?

Can a system have no solution?

How can a tutor check mastery without a full test?

Continue from the Secondary 2 foundation

Official K110 content and reader note

A 60-second check before teaching two unknowns

Substitution: make one variable speak through the other

Elimination: preserve equality while removing one letter

Graphical solutions connect back to Secondary 2 coordinates

86 original simultaneous-equation learning clinics

1. Substitution 1: isolate a variable

2. Substitution 2: isolate a variable

3. Substitution 3: isolate a variable

4. Substitution 4: isolate a variable

5. Substitution 5: isolate a variable

6. Substitution 6: isolate a variable

7. Substitution 7: isolate a variable

8. Substitution 8: isolate a variable

9. Substitution 9: isolate a variable

10. Substitution 10: isolate a variable

11. Substitution 11: isolate a variable

12. Substitution 12: isolate a variable

13. Substitution 13: isolate a variable

14. Substitution 14: isolate a variable

15. Substitution 15: isolate a variable

16. Substitution 16: isolate a variable

17. Substitution 17: isolate a variable

18. Substitution 18: isolate a variable

19. Elimination 1: cancel opposite y-terms

20. Elimination 2: cancel opposite y-terms

21. Elimination 3: cancel opposite y-terms

22. Elimination 4: cancel opposite y-terms

23. Elimination 5: cancel opposite y-terms

24. Elimination 6: cancel opposite y-terms

25. Elimination 7: cancel opposite y-terms

26. Elimination 8: cancel opposite y-terms

27. Elimination 9: cancel opposite y-terms

28. Elimination 10: cancel opposite y-terms

29. Elimination 11: cancel opposite y-terms

30. Elimination 12: cancel opposite y-terms

31. Elimination 13: cancel opposite y-terms

32. Elimination 14: cancel opposite y-terms

33. Elimination 15: cancel opposite y-terms

34. Elimination 16: cancel opposite y-terms