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Additional Mathematics Tutor Singapore | Simultaneous Equations and Quadratic Inequalities

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by

eduKate Asia
Mathematics textbooks stand behind an open notebook containing written working and a graph, with a calculator and pens alongside.

Secondary 3–4 Additional Mathematics · G2/G3 · Linear–quadratic systems, root conditions and inequalities

A student recently told me that they could solve two simultaneous linear equations but felt lost when one equation contained x squared. Nothing particularly magical had changed. Two expressions for the same y-value still have to agree at an intersection. The unfamiliar part was that the resulting equation could have two solutions, a repeated root or no real solution at all.

Quadratic inequalities create a related puzzle. Instead of asking where a curve meets the horizontal axis, they ask where it lies above or below. A student who has learnt to factorise may still miss one interval, include an excluded endpoint or reverse the inequality incorrectly. These errors often survive dozens of routine worksheets because the learner has memorised a picture without understanding why the sign changes.

If you are looking for an Additional Mathematics tutor in Singapore, I want to make those relationships concrete. This guide connects algebra, graphs, discriminants, sign charts and parameter conditions, with explicit solutions and checks. The 2027 SEC G2 Additional Mathematics K232 and G3 K341 syllabuses both contain simultaneous equations by substitution with one equation linear, quadratic inequalities and real-root conditions. The explanations can therefore support either subject level when aligned with the student’s school sequence.

Explore simultaneous equations and inequality methods

Quick answer: how do I solve a linear-quadratic simultaneous system?

Express one variable using the linear equation, substitute that expression into the quadratic equation and solve the resulting quadratic. Each real root gives a candidate coordinate; substitute the x-value back into the linear equation to obtain its matching y-value. Verify every ordered pair in both original equations.

For a quadratic inequality, first identify the real zeros and the leading coefficient. The roots split the number line into intervals. Decide where the quadratic is positive or negative using the graph or an interval test. Include a root only when the inequality permits equality. For parameter questions, translate words such as “two distinct roots” or “tangent” into the correct discriminant condition before calculating.

Why two equations produce intersections

Let y = x + 2 and y = x² − x + 2. Any common point lies on both graphs, so its y-value from the straight line must equal its y-value from the parabola. Thus x + 2 = x² − x + 2, and after rearrangement x² − 2x = 0.

Factorise to x(x − 2) = 0. Therefore x = 0 or x = 2. At x = 0, the line gives y = 2; at x = 2, the line gives y = 4. The two intersection points are (0, 2) and (2, 4).

Check each in the quadratic: at zero, y = 0 − 0 + 2 = 2. At two, y = 4 − 2 + 2 = 4. Both match the line. Writing the pairs separately prevents the common error of mixing x from one root with y from the other.

A repeated root means the line touches

Consider y = 2x + 1 and y = x² − 2x + 5. Set the two outputs equal: 2x + 1 = x² − 2x + 5, which reduces to x² − 4x + 4 = 0. This is (x − 2)² = 0, so x = 2 is a repeated root.

The common y-value is 2(2) + 1 = 5. Thus (2, 5) is the sole contact point, and the line is tangent to the parabola. The repeated factor explains why there is one distinct x-coordinate, not two crossings.

A student should sketch a curve touching a line and compare it with a curve crossing a line twice. The algebraic root count is a graphical statement. Understanding that connection makes discriminant parameter problems easier later.

The discriminant classifies the intersection count

After substituting a line into a quadratic, the intersection equation often has form ax² + bx + c = 0. Its discriminant D = b² − 4ac determines the number of distinct real x-values: D > 0 gives two, D = 0 gives a repeated root, and D < 0 gives none.

This is not an arbitrary table. The quadratic formula includes the term √D. If D is positive, the plus and minus choices differ. If D is zero, they coincide. If D is negative, no real square root exists. The graph interpretation follows directly.

I ask students to write “two distinct real roots” or “one repeated real root” rather than simply “two” or “one”. Precise language helps prevent confusing algebraic multiplicity with the number of physical intersection points.

A parameter can create two possible tangent slopes

Compare the parabola y = x² − 4x + 5 with the family of lines y = mx + 1. Equating gives x² − (m + 4)x + 4 = 0. Tangency requires D = (m + 4)² − 16 = 0, so (m + 4)² = 16.

Taking both square-root possibilities gives m + 4 = 4 or m + 4 = −4. Thus m = 0 or m = −8. There are two tangent lines through the fixed point (0, 1), not merely one.

For m = 0, the contact occurs at x = 2 and y = 1. For m = −8, it occurs at x = −2 and y = 17. Substituting confirms both lie on their respective lines and the parabola. A learner who keeps only the positive square-root branch loses a valid geometric possibility.

Two intersections versus no intersections

Using the same family, two distinct intersections require (m + 4)² − 16 > 0. Factor the expression as m(m + 8) > 0, which holds when m < −8 or m > 0. The lines with slopes outside the interval between the tangent slopes cut the parabola twice.

No real intersection requires m(m + 8) < 0, so −8 < m < 0. The boundary slopes m = −8 and m = 0 produce tangency, because the discriminant equals zero there.

A short number line shows all three cases. Test m = −4 in the middle interval: the line is y = −4x + 1 and the intersection equation becomes x² + 4 = 0, which has no real solutions. This confirms the classification.

A quadratic inequality is a question about sign

Solve x² − 5x + 6 > 0. Factorise it as (x − 2)(x − 3) > 0. The zeros divide the real line at two and three. The product is positive when both factors are negative or both positive, so x < 2 or x > 3.

Between two and three the factors have opposite signs, so the product is negative. At the roots the expression is zero, and those endpoints are excluded by the strict greater-than symbol.

The graph tells the same story: the upward-opening parabola lies above the horizontal axis outside its real roots and below it between them. A student who reports only x > 3 is missing the left outer interval where two negative factors multiply to a positive value.

Non-strict inequalities include the boundary

Change the question to x² − 5x + 6 ≤ 0. We now want the portion on or below the horizontal axis, so 2 ≤ x ≤ 3. The roots are included because zero is permitted by the inequality.

If the sign were < instead, the answer would be 2 < x < 3. The difference is not a new factorisation method; it is a translation of the word “strictly” or the equality bar in the symbol.

I encourage students to circle the inequality sign before manipulating anything and compare the final endpoints with that original sign. Many lost marks arise from forgetting the condition after correct algebra.

A negative leading coefficient reverses the picture

Solve −2x² + 8x − 6 > 0. Factor out −2: −2(x² − 4x + 3) > 0. Dividing both sides by negative two reverses the inequality, so (x − 1)(x − 3) < 0.

The product is negative between the roots, giving 1 < x < 3. Alternatively, sketch the downward-opening parabola with roots one and three; it is above the axis between those values.

A typical mistake is dividing by −2 without changing > to <. Students should explain why a negative multiplier reverses order rather than rely on an unexamined sign-flipping rule.

Repeated roots behave differently on a sign chart

Consider (x − 2)² > 0. A square is non-negative and equals zero only at x = 2, so it is positive for all real x except x = 2. The answer is x < 2 or x > 2.

For (x − 2)² ≥ 0, every real number is permitted. The graph touches the horizontal axis at its repeated root and stays on the same side; it does not cross into negative values.

This distinction matters when learners attempt to memorise sign diagrams. A repeated root does not automatically produce the same sign change as a simple root. The factor’s algebraic structure explains what happens.

An always-positive condition without solving roots

Let q(x) = x² − 6x + k. Completing the square gives q(x) = (x − 3)² + k − 9. It is strictly positive for every real x when k − 9 > 0, or k > 9.

We could also calculate D = 36 − 4k and require D < 0, together with the positive leading coefficient. That gives k > 9 again. At k = 9 the graph touches zero and is merely non-negative, not strictly positive.

Students should learn when completed-square form reveals a bound more directly than calculating roots. The best method is the one that answers the requested condition accurately and efficiently.

A simultaneous equation can involve surds

Solve y = x + 1 and y = x² − 3x + 3. Equating gives x + 1 = x² − 3x + 3, so x² − 4x + 2 = 0. The quadratic formula gives x = 2 ± √2.

Substitute into y = x + 1: the points are (2 + √2, 3 + √2) and (2 − √2, 3 − √2). Keep matching signs within each ordered pair. Exact surds preserve the algebraic relationship more clearly than premature decimal rounding.

Check a root by substituting back into the quadratic intersection equation. A student should also expect two symmetric x-values around x = 2 because the equation’s axis of symmetry is x = 2.

A four-week learning sequence

Week 1: use simple simultaneous linear-quadratic equations and verify every ordered pair. Ask what an intersection means graphically. Week 2: introduce repeated roots, tangency, no-intersection cases and discriminants with parameters.

Week 3: study strict, non-strict and downward-opening quadratic inequalities. Use sign diagrams, test points and verbal explanations of included boundaries. Week 4: mix the methods without worksheet headings and ask the learner to select the first step independently.

The sequence is flexible. A student who still makes bracket or factorisation errors needs targeted repair before proceeding to difficult parameters. A student whose algebra is secure may benefit from more non-routine line–curve questions and alternative checks.

What the tutor should diagnose

When a student finds x-values but reports no y-coordinates, the error may be organisation or interpretation. When they set the discriminant equal to zero in every parameter problem, the issue is recognising the number of roots specified in words.

When they lose one outer interval on a quadratic inequality, the issue is sign reasoning. When they divide an inequality by a negative number without reversing its direction, revisit the inequality principle directly. These are different learning needs and should not all be labelled “weak in A-Math”.

The immutable eduKateSG Clementi Mathematics tutorial reference illustrates our teaching benchmark of close attention, clear explanations and structured practice. It does not imply identical timetables or vacancies across every Secondary 3/4 A-Math programme.

Twenty new linear–quadratic and inequality clinic questions

Try these original tasks before reading their solutions. Each has a distinct first decision, a worked explanation, a common wrong turn and a check. They are not a prediction of SEC paper content or order. Their purpose is to help students identify which mathematical relationship applies when an unfamiliar question is presented without a chapter label.

Clinic 1: Two intersections with a different line

Question: Solve y = 3x + 1 and y = x² − x + 1.

Working: Equate the two outputs: 3x + 1 = x² − x + 1, so x² − 4x = 0. The roots are zero and four. For x = 0, y = 1; for x = 4, y = 13. The intersection points are (0, 1) and (4, 13).

Common mistake: A frequent error is finding the roots correctly but forgetting that the question asks for ordered pairs rather than x-values alone.

Check: Substitute both pairs into the original linear and quadratic equations.

Clinic 2: A system with no real intersection

Question: Do y = x + 1 and y = x² − 3x + 6 meet at any real point?

Working: Equate the expressions to obtain x² − 4x + 5 = 0. The discriminant is 16 − 20 = −4, which is negative. Therefore there are no real x-values where the two y-values agree.

Common mistake: The negative discriminant does not mean the parabola or line fails to exist; it means their graphs do not intersect in the real coordinate plane.

Check: Complete the square: x² − 4x + 5 = (x − 2)² + 1, which can never equal zero.

Clinic 3: Parameter producing tangency

Question: The line y = 2x + k touches y = x² − 2x + 3. Find k and the contact point.

Working: Equate: x² − 4x + 3 − k = 0. Its discriminant is 16 − 4(3 − k) = 4 + 4k. Set it to zero: k = −1. The equation becomes (x − 2)² = 0, so x = 2 and y = 3.

Common mistake: The word ‘touches’ tells us to require a repeated real root, not merely any intersection.

Check: Both original graphs give y = 3 at x = 2.

Clinic 4: Two branches of a squared parameter condition

Question: For x² − 2kx + 9 = 0, find k if there are two distinct real roots.

Working: The discriminant is 4k² − 36. Two distinct roots require 4k² − 36 > 0, so k² > 9. This gives k < −3 or k > 3.

Common mistake: Do not take only the positive square-root branch when solving an inequality involving k². The negative outer region is equally valid.

Check: At k = 4, D > 0; at k = −4, D > 0; at k = 0, D < 0.

Clinic 5: Non-strict negative product region

Question: Solve (x − 1)(x + 3) ≤ 0.

Working: The zeros are one and minus three. The upward-opening quadratic is on or below the axis between its roots, so −3 ≤ x ≤ 1.

Common mistake: A strict inequality would exclude the endpoints, but ≤ allows equality. A student should read the sign before marking the number line.

Check: At x = 0 the product is −3, satisfying the inequality; at either endpoint the product is zero.

Clinic 6: A squared factor cannot be negative

Question: Solve (x − 4)² < 0 over the reals.

Working: Every real square is non-negative, so there is no real solution to the strict inequality. For the changed condition (x − 4)² ≤ 0, the only solution would be x = 4.

Common mistake: Do not invent two intervals merely because the expression is quadratic; its completed-square form already establishes a lower bound.

Check: At x = 4 the value is zero, and everywhere else it is positive.

Clinic 7: Always-positive quadratic without real roots

Question: Show that x² + 4x + 5 > 0 for every real x.

Working: Complete the square: x² + 4x + 5 = (x + 2)² + 1. The square is at least zero, so the whole expression is at least one and is therefore strictly positive.

Common mistake: Do not say D < 0 automatically makes every quadratic positive. The leading coefficient matters.

Check: The discriminant is 16 − 20 = −4, while the positive leading coefficient ensures the graph sits above the axis.

Clinic 8: Negative leading coefficient and endpoints

Question: Solve −(x − 1)(x − 3) ≥ 0.

Working: Multiplying through by −1 reverses the inequality, leaving (x − 1)(x − 3) ≤ 0. The product is non-positive between its roots, so 1 ≤ x ≤ 3.

Common mistake: If the negative multiplier is ignored, the outer intervals may be incorrectly selected.

Check: At x = 2 the original equals one, so the middle interval is correct. At the roots it equals zero and is allowed.

Clinic 9: Non-negative for every x

Question: Determine k if x² + 2x + k ≥ 0 for all real x.

Working: Complete the square: x² + 2x + k = (x + 1)² + k − 1. Its minimum is k − 1 at x = −1. Non-negativity for every real x requires k ≥ 1.

Common mistake: Strict positivity would instead require k > 1. The equality at the boundary changes the final answer.

Check: At k = 1 the expression is (x + 1)², which is never negative and is zero at one point.

Clinic 10: Always-negative quadratic

Question: Find k if −x² + 6x + k is strictly negative for every real x.

Working: Complete the square: −x² + 6x + k = −(x − 3)² + 9 + k. The maximum value is 9 + k. Strict negativity everywhere requires 9 + k < 0, so k < −9.

Common mistake: For a downward-opening parabola, analyse the maximum rather than the minimum. A negative discriminant alone is not enough without the leading-coefficient sign.

Check: At k = −10 the maximum is −1, so every function value is negative.

Clinic 11: Quadratic inequality with surd boundaries

Question: Solve x² − 4x + 1 < 0.

Working: Complete the square: (x − 2)² − 3 < 0, so (x − 2)² < 3. The real interval is 2 − √3 < x < 2 + √3.

Common mistake: There are two boundaries, not simply x < 2 + √3. The left branch matters.

Check: The parabola is upward-opening and has minimum −3 at x = 2, so it lies below zero between two real roots.

Clinic 12: Surd coordinates from simultaneous equations

Question: Solve y = 2x + 3 and y = x² − 4x + 9.

Working: Equating gives x² − 6x + 6 = 0, so x = 3 ± √3. The corresponding y-values are 9 ± 2√3. The points are (3 + √3, 9 + 2√3) and (3 − √3, 9 − 2√3).

Common mistake: Do not mix the positive x branch with the negative y branch when reporting ordered pairs.

Check: Both x-values are symmetric around x = 3, matching the quadratic’s axis of symmetry.

Clinic 13: A fixed intersection plus a parameter-dependent root

Question: Compare y = x² − 2x + 1 with y = mx + 1. Find when their intersections are distinct.

Working: Equating gives x² − (m + 2)x = 0, or x[x − (m + 2)] = 0. One intersection has x = 0; the other has x = m + 2. They are distinct when m ≠ −2.

Common mistake: A student might assume every line meets a parabola twice. At m = −2 the roots coincide at x = 0 and the line is tangent.

Check: Substitute m = −2 to obtain x² = 0, a repeated root.

Clinic 14: Profit-style modelling and a sign interval

Question: A simplified model is P(x) = −x² + 6x − 5. Find real x for which P(x) ≥ 0.

Working: Factor P(x) = −(x − 1)(x − 5). The inequality is equivalent to (x − 1)(x − 5) ≤ 0, so 1 ≤ x ≤ 5. This is a mathematical model and does not establish actual business profit.

Common mistake: The graph opens downward, and the non-negative region lies between the roots. Do not choose the outer intervals from an upward-parabola mnemonic.

Check: At x = 3, P(3) = 4, while at x = 0 it is −5.

Clinic 15: Complementary intervals for the same curve

Question: Solve x² − 6x + 8 ≥ 0 and compare with x² − 6x + 8 < 0.

Working: Factorise as (x − 2)(x − 4). Non-negative values occur for x ≤ 2 or x ≥ 4. Strict negative values occur for 2 < x < 4. These two sets cover the real line without overlap.

Common mistake: Be careful that the equality roots belong only to the non-negative solution set.

Check: At x = 3 the expression is −1, and at x = 1 it equals three.

Clinic 16: Intersections with a horizontal line

Question: For y = x² − 6x + 8, determine when the horizontal line y = k has two distinct real intersections.

Working: Equate x² − 6x + 8 = k. The discriminant is 36 − 4(8 − k) = 4 + 4k. Two distinct real roots require k > −1; tangency occurs at k = −1.

Common mistake: A learner who forgets the vertical position of the vertex may misclassify the threshold.

Check: The completed square y = (x − 3)² − 1 shows the graph’s minimum directly.

Clinic 17: Connected root and turning-point reasoning

Question: The roots of y = x² − 10x + 21 are three and seven. Find the axis and turning-point value.

Working: The axis of symmetry is halfway between the roots at x = 5. Substitute to obtain y = 25 − 50 + 21 = −4. Alternatively, complete the square to (x − 5)² − 4.

Common mistake: A root is an x-value where y = 0; the vertex generally has a non-zero y-coordinate.

Check: At x = 3 and x = 7 the output is zero; symmetry around x = 5 confirms the vertex.

Clinic 18: Finding a circle-line intersection through algebra

Question: Find real intersections of y = x + 1 with x² + y² = 25.

Working: Substitution gives x² + (x + 1)² = 25, or x² + x − 12 = 0. Factorise as (x + 4)(x − 3) = 0. Points are (−4, −3) and (3, 4).

Common mistake: Write both coordinates and verify in the original circle; a root alone is incomplete.

Check: For both points the squared coordinate sum is 25.

Clinic 19: Testing a strict versus non-strict boundary

Question: Determine when x² − 2x + k has exactly one real root, then when it is strictly positive everywhere.

Working: D = (−2)² − 4k = 4 − 4k. Exactly one repeated root requires k = 1. Strict positivity everywhere requires the upward-opening parabola to have minimum k − 1 > 0, hence k > 1.

Common mistake: Do not confuse a repeated root at the boundary with the no-real-root condition for strict positivity.

Check: At k = 1 the expression is (x − 1)², which touches zero.

Clinic 20: A real-world interval constraint

Question: A rectangle has perimeter 28 cm. Let one side be x cm and find its area and its maximum.

Working: The adjacent side is 14 − x, with 0 < x < 14. Its area A = x(14 − x) = −(x − 7)² + 49. The maximum is 49 square centimetres at x = 7.

Common mistake: An algebraic expression may accept real inputs outside the feasible rectangle dimensions; the physical domain rules them out.

Check: The rectangle becomes a 7-by-7 square with perimeter 28 cm.

A learner who can factorise still needs method selection

Topic-labelled drills are useful during first instruction. A worksheet titled “Quadratic Inequalities” gives students a clue about the method before they read the question. Mixed sets remove that clue and require them to infer from the words “positive”, “two intersections” or “touches”. A student who succeeds on drills but not mixed questions may need recognition practice more than extra algebra repetition.

Ask the learner to write one sentence explaining the first method before calculating. When the method is chosen correctly, compare whether execution and checking are also accurate.

Why graph sketches are worth the time

A rough parabola predicts whether a quadratic is above or below zero between roots, whether a tangent has one contact point and where the vertex sits relative to the crossings. A sketch is not a substitute for exact algebra, but it can expose a contradiction before a wrong answer is submitted.

Mark the roots, opening direction and axis of symmetry from known information. If the completed-square minimum says the graph never reaches zero, a claim of two real roots must be investigated.

A four-week revised practice sequence

Week 1 revisits factorisation, signed numbers and exact algebra. Week 2 teaches discriminant conditions through lines and parabolas. Week 3 uses sign diagrams, repeated roots and interval endpoints. Week 4 presents mixed simultaneous, inequality and parameter tasks without method labels.

Do not assume every student should advance at the same pace. If the distributive law is unstable, repair it before increasing difficulty. If the learner already has accurate algebra, spend more time on interpretation and checking.

How to use this clinic without memorising it

Read a question, hide the worked explanation and attempt a solution on blank paper. After marking, write the first invalid line, the underlying reason and a short changed example. Return to the changed example after a few days, not just five minutes later while the model is fresh.

This delayed attempt is a better indicator of retention than recognising yesterday’s solution. It also helps the student avoid using a tutoring session as a sequence of passive demonstrations.

What a useful tutor report might say

A specific report could say: “The student can now solve line–parabola simultaneous equations and correctly pair coordinates, but still misses one outer interval when solving positive quadratic inequalities.” That points directly to the next sign-chart lesson.

The vague statement “needs to work harder” gives no usable teaching direction. Good feedback should tell parents what the learner can already do independently and what exact reasoning still needs support.

What students should check at the end

For a simultaneous system, substitute each ordered pair into both originals. For a quadratic inequality, test a number in each claimed interval and verify endpoint inclusion. For a discriminant condition, check a boundary and a value on either side. For a model, state permitted dimensions and units.

These checks can be carried out with simple arithmetic or sketches and reduce dependence on a tutor declaring the final answer correct.

Frequently asked questions about simultaneous equations and inequalities

Is the discriminant a method for every quadratic question?

No. It is useful for classifying real roots and related tangency or intersection conditions. Completing the square can be clearer for maxima, minima and always-positive properties. Choose the method according to the target.

Does a repeated root count as two intersections?

No. Algebraically the solution has multiplicity two, but the curve and line share one distinct x-coordinate of contact in the standard tangency situation.

Why are there sometimes two answer intervals?

An upward-opening quadratic with two distinct real roots is positive outside both roots. The two outer regions must both be included when the question asks for positivity.

What happens when the leading coefficient is negative?

The parabola opens downward. The sign pattern relative to its roots reverses compared with an upward-opening graph. A test point or careful sign chart avoids a memorised-rule mistake.

Do I include the roots in an inequality answer?

Include them only when equality is permitted by ≤ or ≥. Strict inequalities < and > exclude the zeros because zero does not satisfy strict positivity or negativity.

Can I use the quadratic formula for a simultaneous system?

Yes, after substituting the linear equation you may solve the resulting quadratic by factorisation, completing the square or the quadratic formula. Verify the coordinate pairs in the original system.

Does no real intersection mean the equations are invalid?

No. Each equation may define a perfectly valid real graph. They simply do not share a real point satisfying both equations simultaneously.

Are these topics required in G2 and G3 Additional Mathematics?

Yes, the 2027 K232 and K341 syllabuses both include linear–quadratic simultaneous equations by substitution, quadratic inequalities and discriminant-related root conditions. The overall syllabus levels still differ.

How do I make parameter inequalities easier?

Write the discriminant expression, factor it if practical, mark its zeros on a number line and test the sign in each region. Keep strict or non-strict endpoints consistent with the question.

Will practising these questions guarantee a high exam mark?

No. They support reasoning within important syllabus topics but do not represent the whole assessed course. Success depends on broader preparation, accuracy and independent performance on the actual paper.

Official sources and follow-up lessons

Both the 2027 G2 K232 and 2027 G3 K341 Additional Mathematics syllabuses include simultaneous equations with a linear equation, quadratic inequalities and discriminant conditions. Use the correct examination-year document for the rest of the subject.

For connected study, read quadratic functions, graphs and discriminants, coordinate geometry, circles and tangents, and word problems and constraints. Follow the link that addresses the actual misconception.

Make the graph and algebra agree

Simultaneous equations and quadratic inequalities become less mysterious when a student understands that algebra describes where curves meet and which regions lie above or below the axis. The discriminant counts intersections, factorisation locates roots, and sign analysis gives intervals. Those are related views of the same mathematics.

For Secondary 3/4 Additional Mathematics tutoring, contact eduKate Singapore or enquire on WhatsApp with the child’s recent school work. We teach near Sixth Avenue MRT at 8 Fourth Avenue, Singapore 268674; confirm current class suitability and availability. Properly taught kids shine a bright light into the future.

Choose the next guide according to your current learning need. Paper preparation, algebraic reasoning and trigonometric methods are connected, but they answer different questions. The official subject-level syllabus and your school lessons should still guide the order of revision.