How to survive G1 Mathematics Equations

To survive G1 Mathematics equations, make the letters mean something before moving numbers around. In the 2027 Singapore–Cambridge SEC, G1 Mathematics is syllabus K110. Its Number and Algebra content includes expressions, formulas, linear equations, simple fractional equations, simultaneous linear equations in two variables, and quadratic equations solved using the provided formula. A student who can explain what x represents, preserve equality through each operation and check the answer will often make better progress than someone who memorises ten unrelated tricks.

The practical strategy is identify → model → solve → verify. Identify the unknown and the quantities given. Model their relationship as an equation, rather than guess which operation sounds familiar. Solve step by step. Finally substitute back into the original statement to see whether the result satisfies the problem. This works for a $34 ticket purchase, two unknown quantities in a pair of equations, or a quadratic about a mathematical model. The goal is not advanced algebra for its own sake; it is dependable thinking.

The official source is SEAB’s 2027 G1 Mathematics K110 syllabus. Both Mathematics papers are compulsory, last 90 minutes, carry 50 marks and each contribute 50%. Number and Algebra is shared by both papers. Paper 1 also covers Geometry and Measurement, while Paper 2 covers Statistics and Probability. The provided 2027 formula list includes the quadratic formula and compound interest, and SEAB warns that omission of essential working leads to lost marks. Confirm the correct examination year and school subject level before choosing practice materials.

Algebra opens useful possibilities: reading technical formulas, estimating costs, working with rate and proportion, understanding graphs and explaining relationships in further education. It does not guarantee a specific job or programme, nor does studying Mathematics at G1 close every route. Singapore’s SEC and post-secondary arrangements consider a student’s full subject combination and current programme requirements. A grounded next action is to repair one recurring algebra misunderstanding and prove improvement on a changed example.

Five-minute algebra survival checklist

Write down what each letter means, including its units or whether it represents a whole-number count.
Simplify one side carefully, watching brackets, negative signs and like terms.
Perform an equivalent operation to both sides when solving an equation; do not rely on unexplained transposition.
For two equations, verify that the result satisfies both, not only one.
For a quadratic, identify a, b and c after writing the equation equal to zero; use the provided formula accurately.
Keep the answer within the real context and show essential working, even though an approved calculator may be used.

The official K110 assessment map

Component	2027 specification	Meaning for algebra revision
Paper 1	90 minutes, 50 marks, 50%; Number and Algebra plus Geometry and Measurement	Algebra may appear in short and contextual questions
Paper 2	90 minutes, 50 marks, 50%; Number and Algebra plus Statistics and Probability	Algebra remains examinable in a separate paper
N5 algebraic expressions/formulae	Expressions, evaluating, expanding, factorising, changing subject of simple formulas, patterns	Meaning and operations matter
N6 functions/graphs	Coordinates, linear/quadratic function graphs, gradient, maximum/minimum and symmetry	Interpret graphical evidence
N7 equations	Linear, simple fractional, simultaneous linear and quadratic equations by formula	Set up, solve, check
Examination notes	Approved calculator, provided relevant formulae, essential working; generally 3 significant figures for non-exact values	Show method and apply stated accuracy

A critical boundary: K110 excludes certain extra algebraic identities such as dedicated expansions of (a±b)² and difference of squares, and addition/subtraction of more complex algebraic fractions as separately listed requirements. Yet it includes expansion of two linear expressions and simple x²+px+q factorisation. The right learning plan follows the exact 2027 content rather than forcing G2 or G3 worksheets on a child who still needs stable foundational meaning.

What an equation really says

An equation states that two expressions have the same value. Solving it means finding the value or values that make that statement true. If a ticket cost x dollars and an organiser adds a fixed four-dollar charge to three identical tickets, 3x+4=34 represents a meaningful purchase. Subtracting four from each side and then dividing by three gives x=10. Substitution confirms that 3(10)+4=34.

This is stronger than saying ‘the +4 jumps over and becomes −4’. Equivalent operations preserve equality. The phrase ‘move it across’ can be a shorthand for someone who already understands the procedure, but it becomes a trap when signs, fractions and brackets appear. Rebuild from balance before pursuing speed.

Contents: find the first unstable skill

Algebra meaning · Equations and fractions · Simultaneous equations · Quadratic formulas · Original worked problems · Frequently asked questions

1. Letters represent quantities, not mysterious objects

A letter such as x can stand for an unknown price, distance, number of people or a changing value. It does not automatically mean ‘multiply’. Students need to attach meaning before operating on the symbol. The same x can represent different quantities in unrelated questions.

Imagine a notebook costs x dollars. The expression 3x means three times that price. If a student knows x=4, the total is $12. A variable is a compact way to express relationships, not evidence that a problem is too advanced.

Ask the learner to describe each term in words. When the context changes to four boxes containing x pencils, the multiplication remains, while the units and meaning of x change.

2. Understand algebraic shorthand

The expressions 3x, xy, x² and 3(x+2) contain multiplication even when the multiplication symbol is omitted. A pupil who reads 3x as the number thirty-something may not have internalised algebraic notation.

Use a simple cardboard representation: three bags each hold x counters. The expression 3x describes all counters; x+3 describes one bag’s contents plus three extra. They are not equal for all x.

Practise translating notation to ordinary words, then back to notation. This visible-to-abstract transition is useful before more complex expressions.

3. Difference between expression, equation and formula

An expression is a mathematical quantity written in symbols; an equation asserts equality; a formula describes a relationship among quantities. Students may be told to simplify, solve or substitute, and those instructions require different actions.

For 2x+5, simplification may be possible but no numerical solution for x follows without more information. For 2x+5=17, x can be solved. In C=5+2n, the formula can be used to calculate cost for different item counts.

Ask what is required before calculating. Learning the difference stops students trying to solve expressions which contain no equation.

4. Like terms depend on the variable part

Terms such as 3x and −5x are like terms because they have the same variable component, while 3x and 3x² are not. Combining unlike terms can destroy the meaning of an expression.

The expression 4x+2x−3 becomes 6x−3. The term −3 remains a constant rather than disappearing into the x terms. If x=2, both original and simplified forms equal nine.

Substitution is a powerful verification habit. It checks that simplification has preserved the expression, not merely produced something tidier.

5. Treat negative signs as part of the term

Negative coefficients are easily lost when a pupil skips steps. The sign tells us whether a term adds or subtracts a quantity, and it must travel with that term through legitimate algebraic operations.

For 7−4x+2x, combine −4x and +2x to obtain 7−2x. A pupil who combines 4 and 2 as positive values without the sign would produce a different expression.

Use coloured marking in early guided practice if helpful, but later attempt a changed expression without those supports. Independence is the aim.

6. Expand a bracket fully

Distributive multiplication applies to every term inside a bracket. A pupil who multiplies only the first term or drops a minus sign may get a plausible-looking answer. The meaning is that each item in the bracket contributes to the total.

The expression 3(x+4) becomes 3x+12, while −2(3x−5) becomes −6x+10. Check using a simple value of x; the original and expanded expressions must match.

Do not memorise the phrase ‘change all signs’ for every bracket. The exact multiplier determines the new signs and coefficients.

7. Expand two linear factors accurately

K110 includes the product of two linear expressions. The method multiplies every term in the first factor by every term in the second, then combines appropriate terms.

For (x+2)(x+3), multiply to obtain x²+3x+2x+6, giving x²+5x+6. The constant arises from 2×3, and the middle coefficient from 2+3.

Ask the student to verify at x=1: the product is 3×4=12, matching 1+5+6. Then change one factor and repeat from meaning.

8. Recognise when expression evaluation is needed

Evaluation replaces variables with the values supplied, then performs the indicated operations according to correct precedence. The pupil should not rearrange the task into a different equation or guess a new value for a missing variable.

For 2x²+3x when x=−2, x² is four, so the result is eight minus six, or two. Squaring a negative value within the variable is not the same as merely placing a minus sign before a positive square.

Practise parenthesis carefully on the calculator and check by writing the substitution line explicitly.

9. Use a formula as a relationship

A formula may connect cost, distance, time or another pair of quantities. Understanding the units and roles matters as much as evaluating the arithmetic. A pupil who copies letters without identifying them may misuse a valid formula.

If C=5+2n represents a five-dollar starting fee and two dollars per item, then n=7 gives C=19 dollars. The fixed fee is not the same as the rate per item.

Change the fee or rate and ask how the equation and graph change. This demonstrates generality instead of one remembered total.

10. Change the subject of a simple formula

Changing the subject means rearranging a relationship to express a different quantity, while preserving equality. It is useful when an exam supplies an equation but asks for a variable other than the one isolated in the original.

If C=5+2n, subtract five and divide by two to get n=(C−5)/2. For C=19, n=7. The result is reasonable if n counts items and must be whole under the context.

Students should learn to reverse operations in the right order rather than copy a memorised rearrangement from a similar-looking formula.

11. Solve linear equations one equivalent step at a time

For a linear equation in one unknown, isolate the variable by applying legitimate operations to both sides. The process should remain intelligible so an error can be located. Speed comes from practice after understanding.

For 5x−7=18, add seven to both sides: 5x=25, then divide by five: x=5. Substitute x=5 into the original to verify 25−7=18.

Try a version with a negative coefficient and insist on verification. A correct method must survive the changed sign.

12. Handle variables on both sides

Equations such as 4x+3=2x+15 require collecting variable terms consistently. Students sometimes subtract from one side but forget the other, or change a sign without a corresponding operation.

Subtract 2x from both sides to obtain 2x+3=15, then subtract three and divide by two, giving x=6. Both sides of the original equal twenty-seven.

Ask why it would be wrong simply to combine 4x and 2x as 6x in the starting equation: the terms are on different sides of equality.

13. Solve an equation with a bracket

Brackets can be expanded before isolating the variable, or equivalent operations may be used without expanding when appropriate. Choosing a clear approach and checking signs is more important than using one technique in every problem.

For 3(x−2)=18, divide both sides by three to get x−2=6, so x=8. Expanding first to 3x−6=18 reaches the same result.

Two different valid paths are reassuring when the learner can explain why both preserve equality.

14. Make fractional equations understandable

K110 includes simple fractional equations reducible to linear ones. Denominators represent division, not decorative lines. Multiply through by a suitable nonzero denominator to obtain an equivalent equation while respecting excluded values.

For x/3+(x−2)/4=3, multiply by twelve: 4x+3(x−2)=36. Then 7x−6=36, giving x=6. Substitute back: 2+1=3.

Explain why the twelve multiplies every term, including the right side. This prevents the common error of clearing only one denominator.

15. Check denominator restrictions where relevant

A fractional expression is undefined when its denominator becomes zero. A student should recognise such restrictions, especially if the unknown appears in a denominator, before accepting a proposed solution.

For 3/(x−2)=6, x cannot equal two. Multiplying by (x−2) for permissible values gives 3=6x−12, so x=2.5. This value does not make the denominator zero.

Always substitute the answer back. A value which solves an altered equation but makes the original denominator zero is not valid.

16. Distinguish a solution from a reasonable count

Mathematics may produce a valid numerical answer which does not suit a context. If x represents a number of chairs, a negative fraction is generally not an appropriate practical answer; if x is a temperature, negative values can be meaningful.

An imagined equation for tickets yields x=8; that is a reasonable whole number. Another model yields x=8.5 tickets, suggesting either the context permits fractions or an assumption has been mishandled.

Do not silently round an invalid result to the nearest integer. Explain the constraint and review the modelling.

17. Use real-world wording to build equations

Phrases such as ‘twice the number’, ‘four more than’, ‘shared equally’ and ‘after a fixed charge’ express different algebraic relationships. The learner should name the unknown, then translate the full relationship rather than search for one trigger word.

A fictional class purchases three identical tickets and pays a fixed $4 handling fee, for a total of $34. The equation is 3x+4=34, not 3(x+4)=34.

Ask what the brackets would mean in the second expression: a separate four-dollar fee added to every ticket, which changes the scenario.

18. Check an answer by substitution

Substitution asks whether the solved value makes every original statement true. It can catch sign errors, an algebraic fraction mishandled through multiplication, or a misplaced constant.

For 2x−5=9, x=7 satisfies fourteen minus five equals nine. If a learner reports x=2, substitution immediately shows that the left side is minus one.

Encourage an independent check as a normal part of the method rather than something used only when the answer key disagrees.

19. Interpret the graph of a linear relationship

The graph of y=ax+b is a straight line, with gradient a and vertical intercept b in the standard coordinate axes. The quantities and units define the meaning of those features.

In a fictional service C=3n+5, the gradient is three dollars per item and the intercept a five-dollar fixed charge. A line with a negative gradient might describe a value decreasing as the horizontal quantity increases.

A candidate should not call every gradient speed. Look at the actual axis labels and units.

20. Understand simultaneous equations as two constraints

A pair of simultaneous linear equations represents two relationships that must hold at once. Solving only one equation leaves infinitely many possible pairs; the second constraint helps determine their intersection.

Suppose x+y=12 and x−y=4. Both need to be true. Adding gives 2x=16, so x=8 and y=4. These values satisfy each relation.

Ask students to explain why x=12, y=0 is not a valid pair even though it satisfies the first equation. The second is essential.

21. Use substitution when one variable is easy to isolate

The substitution method expresses one unknown through the other and replaces it in the second equation. This creates an equation in one variable. It is particularly clear when one original equation already has a variable alone.

For x+y=10 and x=2y+1, substitute: (2y+1)+y=10, so 3y=9, y=3 and x=7. Check both statements.

The aim is not to follow one memorised ‘substitution question’ shape but to recognise the advantage of an already isolated variable.

22. Use elimination to remove one variable

Elimination adds or subtracts equivalent equations so that one variable cancels. Multiplication of a whole equation may be needed first. A student should keep every term, including constants, when transforming.

For 2x+y=11 and x−y=1, adding gives 3x=12, so x=4; substitute to find y=3. The two equations are both satisfied.

Ask why multiplying only the x term by two but not the other terms would be invalid. Equivalent whole equations preserve the original constraints.

23. Match coefficients correctly before elimination

When coefficients do not initially cancel, the learner can multiply an entire equation by a suitable factor. Sign errors often occur when subtracting equations, especially if brackets are not used.

Consider 2x+3y=16 and 3x+2y=19. Multiply the first by three and the second by two to get 6x+9y=48 and 6x+4y=38. Subtract to obtain 5y=10, giving y=2 and x=5.

Check the pair in both originals: 10+6=16 and 15+4=19. A line-by-line record makes the method inspectable.

24. Understand the graphical solution

Two linear equations in two variables can be represented as straight lines. A point where the lines intersect satisfies both equations. If the drawing is approximate, the coordinates read off may be estimates requiring verification.

For x+y=10 and x−y=2, the lines meet at (6,4). Reading that point explains the numerical solution visually.

Students should not invent exact integer coordinates from a roughly plotted intersection; calculate if precision is requested and the necessary data are available.

25. Distinguish one, none and infinitely many pairs

Depending on their relationships, two lines can intersect at one point, be parallel and distinct, or coincide. The graphical interpretation helps explain why a simultaneous system need not always have exactly one distinct solution.

In a simple model, x+y=4 and 2x+2y=8 describe the same line; every point on it satisfies both. By contrast, x+y=4 and x+y=7 cannot both hold.

These are useful enrichment explanations, not an invitation to invent extra advanced examination formats outside the K110 specification.

26. Form two equations from a purchase story

Two unknown item prices can be described by two different purchase totals. A correct setup must identify what each quantity counts and whether the same unit price applies in both conditions.

An imagined purchase has two drinks and one sandwich totalling $11, while one drink and one sandwich cost $7. Let d be a drink and s a sandwich. The equations are 2d+s=11 and d+s=7.

Subtract to find d=4 and s=3. Check both purchases. If a student cannot explain what d and s mean, their algebraic solution remains fragile.

27. Use linear graphs to validate a context

A graph can show a fixed fee and per-unit rate or two competing pricing schemes. Intersections can represent a condition at which two modelled costs are equal, but only within the meaningful input range.

Suppose Plan A costs 5+3n and Plan B costs 9+2n dollars. They are equal when 5+3n=9+2n, giving n=4. The common modelled cost is $17.

At another count, compare both costs separately. Do not claim the plans are the same for every n merely because they cross once.

28. Recognise what makes an equation quadratic

A quadratic equation in one variable has a nonzero squared term as its highest relevant power after simplification. The K110 syllabus includes solving such equations with the provided quadratic formula. Students should not automatically apply linear-equation steps to x² terms.

For x²−5x+6=0, the squared variable means x cannot be isolated by one ordinary division alone. There may be two roots which must be considered.

Before calculating, write the equation as ax²+bx+c=0 and identify all three coefficients including their signs.

29. Identify a, b and c with signs intact

In ax²+bx+c=0, a is the x² coefficient, b the x coefficient and c the constant. Forgetting a negative sign can produce a result that looks plausible but fails the equation.

For 2x²−7x+3=0, the coefficients are a=2, b=−7, c=3. If the student enters b=7, the sign of −b and the discriminant terms are mishandled.

Write a substitution line into the formula rather than enter all numbers into the calculator from memory.

30. Apply the provided quadratic formula

The official formula is x=(−b±√(b²−4ac))/(2a) for ax²+bx+c=0. It uses both plus and minus, which may give two solutions; the square-root expression and denominator must be grouped correctly.

For x²−5x+6=0, a=1,b=−5,c=6. The discriminant is 25−24=1; x=(5±1)/2 gives x=3 or x=2. Both satisfy the equation.

Use parentheses on the calculator and show each step. A single missing bracket can change division by 2a into an entirely different computation.

31. Understand the discriminant as a check

The expression b²−4ac inside the square root determines whether the quadratic formula gives real solutions in the model. For G1 revision, the important skill is evaluating the expression correctly and interpreting the resulting roots in context.

A negative discriminant would mean there is no real-number solution under the usual formula interpretation. Students should not invent a real root from the square root of a negative number when the question expects real values.

The official task may provide sufficient information to interpret a graph or context; avoid going beyond the K110 learning outcomes by demanding unrelated advanced complex-number techniques.

32. Verify both quadratic roots

An equation can have more than one candidate value, and each must satisfy the original. Students who calculate both branches but report only the first may answer incompletely unless the context restricts which root matters.

For x²−5x+6=0, 2 gives 4−10+6=0 and 3 gives 9−15+6=0. Both are valid mathematical solutions.

If x represents a length, negative roots may be impossible physically. State that contextual restriction rather than deleting an answer without explanation.

33. Factorise a simple monic quadratic

The syllabus also includes factorisation of quadratic expressions x²+px+q. Factorisation rewrites a polynomial as a product and can help verify roots or simplify an expression, though K110 specifically lists the quadratic formula for solving equations.

For x²+5x+6, two numbers with sum five and product six are two and three, so it factorises to (x+2)(x+3). Expanding confirms equivalence.

Practise reversing the process using a new constant and coefficient. The learner should understand both sum and product rather than memorise pairs.

34. Use graph shape to think about roots

A quadratic function produces a parabolic graph whose leading coefficient affects whether it opens up or down. Roots correspond to x-values at which the graph meets the x-axis when such intersections exist.

For y=x²−5x+6, the graph crosses the x-axis at 2 and 3, matching the equation’s roots. A candidate can link algebraic and graphical forms without claiming that a sketch yields perfect numerical precision.

The syllabus includes quadratic graphs, maximum/minimum points and symmetry; graph interpretation and algebra should support each other.

35. Recognise minimum and maximum on a parabola

A parabola opening upward has a lowest point, while one opening downward has a highest point. In a real-world model, the corresponding optimum may be meaningful only within the allowed input range.

For y=(x−2)²+1, the model has minimum value one at x=2. The shifted square makes that obvious, but the student should still recognise the axes and domain.

This is a conceptual practice exercise, not an invitation to learn calculus or assume a fictional mathematical optimum is a safe engineering recommendation.

36. Keep coordinates in the right order

An ordered pair (x,y) states the horizontal coordinate first and vertical coordinate second. Reversing them gives a different point and can invalidate a graphical solution.

The point (3,7) is not generally the same as (7,3). Plotting both on simple squared paper makes the difference visible.

Tie axes to meaningful quantities, such as number of items and total price, so the coordinate convention becomes more than a symbol rule.

37. Check whether the units suit the answer

Equation solutions often describe prices, durations, distances or counts. A number without its meaning may be incomplete. A price of ten dollars, for example, is not ten metres.

For a fictional ticket problem, x=$10 answers a unit-price question. If the question asks the combined cost of three tickets, the answer is $30 before any separately stated fee.

Students should write the final statement in context and check magnitude. This habit prevents confusing intermediate quantities with the result requested.

38. Keep the calculator as a checking tool

K110 permits approved calculators in both papers, but entering a full expression correctly is only one part of problem solving. The child must choose the right equation, identify coefficients and display essential working.

For a quadratic, write the discriminant and numerator clearly before entering the square-root expression. Then verify roots with substitution.

Never use a calculator result to hide a modelling error. An impossible negative count should prompt a context check, not immediate unexamined rounding.

39. Work methodically under a ninety-minute deadline

Both K110 papers contain eleven to thirteen shorter questions and two longer context-based items. A pupil should not spend an excessive part of the session rewriting one solution for cosmetic perfection while other questions remain unanswered.

Practise a personal pacing method based on school-style timed work. Note whether the most frequent delay is interpreting a sentence, clearing a fraction or entering a formula.

Focus future practice on the actual bottleneck, with a short verification step reserved where time allows.

40. Use a six-week repair sequence

Week one stabilises algebraic notation and like terms. Week two develops brackets and expressions. Week three tackles linear and fractional equations. Week four covers simultaneous equations by two methods. Week five practises the quadratic formula and checking. Week six integrates graph interpretation and real-world scenarios.

This is a suggested sequence, not an official timeline or a grade guarantee. Move more slowly if essential fraction or negative-number skills are still uncertain.

The meaningful indicator is that the pupil can complete an unfamiliar equation and explain each step without an adult supplying hints.

41. Understand how families and tutors can help

A parent can ask what x represents and whether the answer satisfies the original story, even without solving the problem for the child. A tutor can diagnose the earliest mistaken operation, rebuild the concept visibly and withdraw support gradually.

A premium three-student small-group lesson near Sixth Avenue can provide close feedback on working and misconceptions, but a lesson cannot guarantee a grade or admissions outcome. School work and independent practice remain important.

The most useful support is a calm correction followed by a changed question. Memorising the tutor’s answer is not the same as understanding.

42. Keep progression and future pathways realistic

Equations and functions support later technical learning, formulas in applied Science, service-cost models, design measurement and many other everyday problems. Strong foundations are valuable whether a student later chooses a technical or service-oriented course.

K110 is a G1 subject, and later subject-level progression and post-secondary pathways depend on school decisions, full SEC results and institution rules. One difficult quadratic problem does not decide a teenager’s future.

A good immediate plan is to identify one unstable skill and practise enough to make the next unfamiliar example manageable.

Six original worked K110 algebra examples

A real-world linear equation

Fictional service: three identical tickets plus a $4 fee cost $34. Let t be the price of one ticket: 3t+4=34. Subtract four to get 3t=30 and divide to get t=$10. Check that three ten-dollar tickets plus four dollars total thirty-four.

Change the fee to $7 and the total to $37. The price is still ten dollars, but the equations and arithmetic steps must be updated. This shows the relationship is understood rather than copied.

An equation with a variable denominator

Original classroom problem: 3/(x−2)=6, x≠2. Multiply both sides by x−2: 3=6x−12, so 6x=15 and x=2.5. The denominator at this value is 0.5, making the original quotient six.

Check why x=2 is excluded even if another algebraic manipulation appears to produce it. Denominator restrictions are part of the original problem, not optional decoration.

Solving two ticket prices

Invented purchase: two adult tickets and one child ticket cost $17; one adult and one child cost $10. Let a and c be ticket prices. Then 2a+c=17 and a+c=10. Subtracting gives a=7, so c=3.

Check both totals: fourteen plus three is seventeen, and seven plus three is ten. The same pair must satisfy both constraints.

A simultaneous equation by elimination

Take 2x+3y=16 and 3x+2y=19. Multiply first by three: 6x+9y=48; second by two: 6x+4y=38. Subtract to find 5y=10, so y=2. Then 2x+6=16, x=5.

Checking 3×5+2×2=19 confirms the other equation. A learner should show where each coefficient came from instead of presenting the final pair mysteriously.

Using the quadratic formula

For x²−5x+6=0, set a=1,b=−5,c=6. The discriminant is b²−4ac=25−24=1. Hence x=[5±√1]/2, giving x=2 or x=3. Direct substitution confirms both roots.

Change the constant to 4. The roots change, so copying two and three cannot be justified. Reidentify the coefficients and evaluate the new discriminant.

An intersection of linear models

Hypothetical plan A costs 5+3n dollars and plan B costs 9+2n, where n is a whole number of items. Equal cost occurs when 5+3n=9+2n, giving n=4 and each total $17.

At n=2, A costs $11 and B $13; at n=6, A costs $23 and B $21. The intersection describes equality at one input, not a universal claim that both plans are the same.

Common mistakes to recognise before examinations

1. The symbol 3x is read as x+3

Explain the multiplication represented by a coefficient and test x=4 to see why the two expressions differ.