Algebra - Solving equations and inequalities: solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph (Higher tier)

Overview

Simultaneous equations ask you to find values of two unknowns that satisfy two conditions at the same time. Think of each equation as a rule: a successful answer must obey both rules, not just one. In a graph, the solution is an intersection because that point belongs to both graphs. In algebra, you reduce the two equations to one unknown, solve it, then recover the matching value of the other unknown.

For Pearson Edexcel GCSE Mathematics, specification 1MA1, A19, candidates must solve simultaneous equations algebraically and find approximate solutions graphically. Linear/linear work is assessed at both tiers; linear/quadratic work is Higher-only. This guide develops both so you can choose a method rather than apply a memorised routine blindly. 1

Questions may say “solve algebraically”, ask you to draw or use graphs, or hide the equations inside a geometry or modelling problem. The supporting skills are rearranging formulae, expanding brackets, factorising, using the quadratic formula and reading coordinates. Clear working makes your method visible. A complete answer usually means an ordered pair, or two ordered pairs, with each x securely matched to its y.

Key Concepts

1. What counts as a simultaneous solution?

The pair (3, 1) solves 2x + y = 7 and x − y = 2 because 2(3) + 1 = 7 and 3 − 1 = 2. Checking only one equation is insufficient: infinitely many points satisfy either straight-line equation individually. Solving simultaneously finds the shared point or points.

An ordered pair is written (x, y), in that order. If x = 3 and y = 1, the point is (3, 1), not (1, 3). In pure algebra, coordinates do not need units. In a word problem, attach the units and interpret the answer using the original definitions of the variables.

Cover and recall: What two substitutions would prove that (3, 1) is a solution? Write both checks before looking back.

2. Two linear equations: eliminate one variable

A linear equation has variables to the first power, such as 2x + y = 7. Elimination combines equations so one variable cancels. Add 2x + y = 7 and x − y = 2 to obtain 3x = 9. Hence x = 3. Substitute into x − y = 2 to obtain y = 1.

Why is this valid? The left and right sides of each equation are equal. Adding equal quantities preserves equality. Multiplying every term by the same non-zero number also preserves its solutions.

If the coefficients do not already match, create matching coefficients. For 3x + 2y = 12 and 2x − 3y = −5, multiply the first equation by 3 and the second by 2. This gives 9x + 6y = 36 and 4x − 6y = −10. Adding gives 13x = 26, so x = 2 and y = 3. Multiply every term, including the constant. Opposite coefficients cancel by addition; equal coefficients cancel by subtraction. Substitution is also valid for linear pairs.

3. Linear/quadratic pairs: substitution is the reliable route

A quadratic equation includes a squared variable, such as y = x² − 3 or x² + y² = 10. Substitution uses the linear equation to replace one variable throughout the quadratic equation. Choose the variable already isolated, or rearrange the linear equation to minimise fractions.

For y = 2x and y = x² − 3, replace y in the quadratic with 2x: 2x = x² − 3. This leaves only x. Rearrange to x² − 2x − 3 = 0, factorise to (x − 3)(x + 1) = 0, then obtain x = 3 or x = −1.

Important correction: elimination is not universally impossible when one equation is quadratic. If both equations give y, subtracting eliminates y and is equivalent to equating the two expressions. However, ordinary linear elimination does not generally remove squared terms in relations such as x² + y² = 10. Substitution is the dependable GCSE method, not a rule that forbids other correct algebra.

Why does this work? At a solution, the linear expression and the replaced variable have exactly the same value. Replacing equals with equals therefore preserves every simultaneous solution.

4. Substitute the whole expression, then expand carefully

Suppose x + y = 4 and x² + y² = 10. Rearrange to y = 4 − x. Since y² means y multiplied by itself, substitute (4 − x)², not 4 − x². Now x² + (4 − x)² = 10.

Expand (4 − x)(4 − x) to obtain 16 − 8x + x². The middle term comes from two cross-products: −4x and −4x. Collecting terms gives 2x² − 8x + 6 = 0. Divide every term by 2 to simplify before factorising.

Similarly, (x + 4)² = x² + 8x + 16, not x² + 16. Candidates commonly lose accuracy by omitting the middle term. Write a squared bracket as two identical brackets if you are uncertain. A negative number squared is positive when the whole negative number is squared: (−3)² = 9, whereas −3² means −9 under the usual order of operations.

5. Solve the resulting quadratic without losing roots

First arrange the equation in the form ax² + bx + c = 0. Factorising works when suitable factors are easy to find. From (x − 3)(x + 1) = 0, at least one factor must be zero, giving two branches: x − 3 = 0 or x + 1 = 0.

If factorising is awkward, use the quadratic formula, which appears on Pearson's Higher-tier formula sheet. Identify a, b and c with their signs. In x² − x − 3 = 0, a = 1, b = −1 and c = −3. The formula gives x = (1 ± √13)/2. Use both signs. Completing the square is another valid Higher-tier method. 1 2

Do not divide an equation by x, x − 1 or another variable expression without checking whether that expression could be zero. Doing so can remove a valid root. Dividing every term by a known non-zero constant, such as 2, is safe. Keep exact fractions or surds unless the question requests rounding. On calculator papers, retain unrounded roots while finding y.

6. Match every x to its own y

For y = 2x, the roots x = 3 and x = −1 give y = 6 and y = −2 respectively. Therefore the answers are (3, 6) and (−1, −2). Four disconnected numbers are a poor finish because their pairing can be ambiguous.

Write each branch on a separate line: “When x = 3, y = 6”; “when x = −1, y = −2”. Then check both equations for each pair if time permits. For (−1, −2), −2 = 2(−1) and −2 = (−1)² − 3. This checks the signs as well as the arithmetic.

Pearson's June 2022 Higher Paper 3, question 22, awarded three method marks followed by one accuracy mark for each correct solution pair. That published five-mark allocation demonstrates why finding x alone is not a complete response. It is not a universal allocation for every question. All original questions in this guide have illustrative, not official Pearson, mark schemes. 3

7. Graphical solutions: intersections, not intercepts

The purple parabola y = x² − 3 and dashed line y = 2x intersect at (−1, −2) and (3, 6). These are the same solutions obtained algebraically. An x-intercept of the parabola solves y = 0; it does not solve the simultaneous pair unless the other equation is also satisfied there.

When drawing graphs, check both axis scales first. Use a ruler for the line. Two accurate points define a straight line, but a third helps catch a plotting error. For the quadratic, calculate enough points to show its shape and turning region, plot them, then draw a smooth curve rather than straight segments. Negative x values must be substituted using brackets.

Read x and y from each intersection using light horizontal and vertical guide lines. Use the accuracy specified in the question. Otherwise report only the precision supported by the grid: a nearest-tenth estimate is often sensible on the practice grid here, but it is not a universal requirement. Do not invent extra decimal places. The exact acceptance interval is determined by the particular mark scheme.

Try before checking: Use this graph to estimate the simultaneous solutions of y = x² − 2 and y = x + 1. Expect two coordinate pairs, approximately (−1.3, −0.3) and (2.3, 3.3). The worked example explains the reading method. A correct algebraic result does not replace a required graphical method.

8. Two, one or no real solutions

For a non-vertical straight line and a genuine parabola y = ax² + bx + c, with a ≠ 0, substitution produces a quadratic. Two distinct real roots mean two intersections. A repeated root means one intersection where the line is tangent. No real roots mean the graphs do not meet.

For y = x² and y = 2x − 1, equating gives x² − 2x + 1 = 0, or (x − 1)² = 0. There is one distinct pair, (1, 1), not two different points. For y = x² and y = −1, x² = −1 has no real solution because a real square cannot be negative.

The discriminant D = b² − 4ac explains these cases: D > 0 gives two real roots, D = 0 a repeated root, and D < 0 no real roots. This is useful enrichment derived from the quadratic formula. Do not assume formal discriminant terminology is required by every GCSE question. One visible intersection on a restricted grid does not prove tangency: a second might lie outside the window.

Mathematical/Scientific Relationships

The following are the relevant relationships for this topic, not a list of every formula on the GCSE course.

Given on formula sheet: x = [−b ± √(b² − 4ac)]/(2a), for ax² + bx + c = 0 and a ≠ 0. Here a, b and c are coefficients including their signs; x is the unknown. Apply the square root to the entire discriminant and divide the entire numerator by 2a. Pearson's 2026 Higher-tier aid supplies this formula; practise with the aid for your examination series. Pearson confirms continued formula-sheet provision for the current specification. 2 4

Must memorise: (u + v)² = u² + 2uv + v² and (u − v)² = u² − 2uv + v². These identities explain squared-bracket substitution. Also know u² − v² = (u − v)(u + v) when difference-of-squares factorisation appears.

Must memorise: PQ = 0 implies P = 0 or Q = 0. This is the zero-product rule. A quadratic must be set equal to zero before applying it to its factors.

Must memorise: y = mx + c describes a straight line: m is its gradient and c its y-intercept. At an intersection of y = f(x) and y = g(x), f(x) = g(x); substitute the resulting x into either expression to find y.

Must memorise: (x − h)² = k gives x = h ± √k for k ≥ 0. This is the square-root step after completing the square. The two signs still matter.

Must memorise if using this optional enrichment: D = b² − 4ac and its root-count interpretation. The expression appears inside the supplied quadratic formula, but the separate interpretation is not supplied on the formula sheet.

For the circle extension, x² + y² = r² describes a circle centred at the origin, where r is the radius. Must memorise as the coordinate-circle relationship. Its underlying Pythagorean relationship a² + b² = c² is Given on formula sheet. Do not confuse the coefficient c in a quadratic with a side labelled c in Pythagoras' theorem. 1 2

Practical Applications

Two pricing conditions can determine two unknown ticket prices. A rectangle with a known perimeter and area gives a linear equation from its perimeter and a quadratic relationship from its area. For side lengths x and y, 2x + 2y = P and xy = A; substitute y = P/2 − x to obtain one quadratic. Reject negative lengths only because the context demands positive measurements, not because negative roots are inherently wrong.

Circle-and-line intersections connect simultaneous equations to coordinate geometry and Pythagoras. Graphs can also show where two models predict the same value. State what each coordinate means in context.

There is no required practical for this mathematics topic. No special unit conversion is built into A19. In contextual questions, make units consistent before forming equations: use either pounds or pence throughout, and one length unit throughout. Area conversions are squared: 1 m² = 10,000 cm², not 100 cm².

Listen, Retrieve and Revisit

The ten-minute female-narrated episode follows this sequence: introduction 0:00–1:00; core concepts 1:00–6:00; exam tips 6:00–8:00; recall quiz 8:00–9:00; summary 9:00–10:00. Listen with a pen and attempt each prompt before hearing the answer. Afterwards, cover the method and solve a fresh problem. Revisit tomorrow and again several days later, mixing algebraic and graphical tasks so you must select the approach yourself.

References