AQA A-level Mathematics: Solving Simultaneous Equations by Elimination and Substitution (7357) – Revision Notes

These notes condense statement B4 of Section B (Algebra and functions) in the AQA A-level Mathematics specification, first teaching 2017. For full explanations and longer worked examples, read the study guide. When you are ready to test yourself, use the practice set. Course links: AQA A-level Mathematics hub, printable checklist and the free 10-minute diagnostics.

The B4 statement in one line

Use elimination and substitution on pairs of equations in two unknowns, including one linear and one quadratic. Both methods must be familiar. You choose whichever suits the pair.

Key ideas

Idea	Summary
Solution	A pair of values (x, y) that satisfies both equations
Graph meaning	A point lying on both graphs
Two lines	One point, no point (parallel) or every point (same line)
Line and quadratic curve	Two points, one point (tangent) or none
Answer format	Matched pairs, exact where the question asks for exact

Method boxes

Elimination (two linear equations)

1. Label the equations (1) and (2).
2. Scale one or both so one letter has equal coefficients.
3. Add if the signs differ, subtract if they match.
4. Solve for the remaining letter.
5. Substitute back to find the other letter.
6. Check in the equation you did not use for step 5.

Substitution (linear and quadratic)

1. Rearrange the linear equation so one letter stands alone.
2. Replace that letter in the quadratic, with brackets.
3. Expand and gather terms as ax² + bx + c = 0.
4. Solve by factorising, completing the square or the formula.
5. Return to the linear equation for each matching value.
6. Write the answers as pairs.

Discriminant cases for a line and a curve

After equating, you have ax² + bx + c = 0 in x. Then:

b² − 4ac	Meaning
positive	line cuts the curve at two points
zero	line is a tangent: one repeated root
negative	line does not meet the curve

If the line contains a letter such as k, the discriminant is an expression in k. Set it equal to zero, greater than zero or less than zero to answer the question. The tangent case also gives the point of contact: the repeated root is x = −b/(2a).

Worked reminders

Elimination. Solve 2x + 5y = 16 and 3x − 2y = 5.

(1)×2: 4x + 10y = 32
(2)×5: 15x − 10y = 25
Add:   19x = 57, so x = 3
From (1): 6 + 5y = 16, so y = 2
Check (2): 9 − 4 = 5 ✓

Quadratic term in xy. Solve x − y = 3 and xy = 28.

x = y + 3
(y + 3)y = 28
y² + 3y − 28 = 0
(y + 7)(y − 4) = 0
y = −7 or y = 4
x = −4 or x = 7

The solutions are (−4, −7) and (7, 4). Both pairs satisfy xy = 28: (−4)(−7) = 28 and 7 × 4 = 28.

Tangent condition with a letter. Find k so that y = 4x + k touches y = x² + 2x + 6.

x² + 2x + 6 = 4x + k
x² − 2x + (6 − k) = 0
Discriminant: 4 − 4(6 − k) = 4k − 20
Tangent: 4k − 20 = 0, so k = 5
Contact: x² − 2x + 1 = 0, x = 1, y = 4(1) + 5 = 9

For two intersections you would need 4k − 20 > 0, so k > 5. For none, k < 5.

Context pair. A rectangle has perimeter 34 cm and area 60 cm². With length x and width y:

x + y = 17 and xy = 60
x(17 − x) = 60
x² − 17x + 60 = 0
(x − 5)(x − 12) = 0

The sides are 5 cm and 12 cm. Define your letters, state the equations, and finish with a sentence that answers the question.

Special cases for two linear equations

What elimination gives	Meaning
A single value for x, then y	One solution: the lines cross
A false statement such as 0 = 7	No solution: parallel lines
A true statement such as 0 = 0	Infinitely many solutions: the same line

Compare the ratios of coefficients before you start. If the x and y coefficients are in the same ratio but the constants are not, the lines are parallel.

Checking your answers

• Put both values back into the equation that you did not use last.
• For a line and curve, check that each point lies on the line and on the curve.
• Check that the number of solutions agrees with the discriminant.
• In a context problem, reject values that cannot happen, such as a negative length.

Link with coordinate geometry

Finding where a line meets a curve is the algebra behind many coordinate geometry questions. Once you have the two intersection points, you can find the midpoint with ((x₁ + x₂)/2, (y₁ + y₂)/2) and the distance with √((x₂ − x₁)² + (y₂ − y₁)²). For a line y = x + c meeting a curve, the two x-values differ by the square root of the discriminant divided by a, which gives a short route to a chord length. See coordinate geometry in the plane for the geometry.

Must-know distinctions

• Elimination versus substitution. Elimination removes a letter by adding or subtracting scaled equations. Substitution replaces a letter by an expression. With a quadratic in the pair, use substitution.
• Equating versus substituting. If both equations start y = …, setting the right-hand sides equal is a quick substitution.
• No solution versus no real solution. Parallel lines give no solution. A line and curve with a negative discriminant give no real intersection.
• Tangent versus intersection. A tangent gives one repeated root, so one point, not two.
• x-values versus points. The quadratic in x gives only x-values. The question wants points.
• Exact versus decimal. Surd answers stay as surds unless you are asked for a rounded form.

Quick self-test

Attempt every question before checking the answers below.

1. Solve 3x + 2y = −1 and 5x − 4y = 13.
2. Solve x + y = 9 and y = x² − 3.
3. Solve x + y = 10 and x² + y² = 52.
4. Find the values of k for which the line y = 2x + k does not meet the curve y = x² + 1.
5. Does the pair 3x − 6y = 2 and x − 2y = 5 have a solution? Explain.
6. Does the line y = 3x − 4 meet the curve y = x²? Show your reasoning.
7. Solve y = x − 1 and y = x² − 2x − 2, giving exact answers.
8. The line y = 4 − x meets the curve x² + y² = 10. Find the points of intersection.
9. The line y = mx, with m > 0, is a tangent to y = x² − 4x + 9. Find m and the point of contact.
10. Write down the quadratic you would solve first to find where y = 5x − 6 meets y = x² − x + 3.

Answers

1. Double (1) and add to (2): 6x + 4y = −2, plus 5x − 4y = 13 gives 11x = 11, so x = 1. Then y = −2. (1, −2)
2. x + x² − 3 = 9, so x² + x − 12 = 0, (x + 4)(x − 3) = 0. (3, 6) and (−4, 13)
3. y = 10 − x gives x² + (10 − x)² = 52, so x² − 10x + 24 = 0. (4, 6) and (6, 4)
4. x² + 1 = 2x + k gives x² − 2x + (1 − k) = 0. The discriminant is 4k. No meeting needs 4k < 0, so k < 0.
5. Multiply the second equation by 3: 3x − 6y = 15. This contradicts 3x − 6y = 2. No solution: the lines are parallel.
6. x² = 3x − 4 gives x² − 3x + 4 = 0, with discriminant 9 − 16 = −7. No, the line does not meet the curve.
7. x − 1 = x² − 2x − 2 gives x² − 3x − 1 = 0, so x = (3 ± √13)/2 and y = x − 1. ((3 + √13)/2, (1 + √13)/2) and ((3 − √13)/2, (1 − √13)/2)
8. x² + (4 − x)² = 10 gives x² − 4x + 3 = 0. (1, 3) and (3, 1)
9. x² − (4 + m)x + 9 = 0 needs (4 + m)² = 36, so m = 2 (rejecting m = −10). The repeated root is x = 3, y = 6. m = 2, point (3, 6)
10. 5x − 6 = x² − x + 3 gives x² − 6x + 9 = 0, which is (x − 3)² = 0, so the line is a tangent.

Exam technique

• Show the substitution line and the rearranged quadratic. They make your method easy to follow and to credit.
• If a question says “show that”, reach the printed result with every line of working.
• Read the command words: “solve” wants points, “find the range of k” wants an inequality in k.
• Keep exact fractions and surds in the working and round only at the end.

Where marks are usually lost

• Scaling one side only, so the equation is no longer balanced.
• Adding when the signs match, or subtracting when they differ.
• Dropping the brackets around a substituted expression before squaring.
• Giving x-values only and not the partner y-values.
• Pairing the wrong y with an x because the substitution went into the quadratic.
• Writing a discriminant with a sign error when b is negative.
• Setting the discriminant greater than zero when the question says “tangent”.
• Rounding too early on surd questions.
• Cancelling a common factor of x, which loses the solution x = 0.

Official syllabus

AQA A-level Mathematics (7357) specification, first teaching 2017, published by AQA. Statement B4, Section B: Algebra and functions.