What you'll learn
This topic combines algebraic substitution with quadratic solving techniques to find two pairs of coordinate solutions where a straight line intersects a curve. Questions on simultaneous equations involving one linear and one quadratic regularly appear in AQA GCSE Mathematics Paper 2 (calculator) and Paper 3 (non-calculator), typically worth 4-6 marks. You'll develop a methodical approach to eliminate one variable, solve the resulting quadratic equation, then find corresponding values for both variables.
Key terms and definitions
Simultaneous equations — two or more equations that must be solved together to find values that satisfy all equations at the same time.
Linear equation — an equation containing variables with power 1 only, forming a straight line when graphed (e.g., y = 2x + 3 or 3x + y = 7).
Quadratic equation — an equation where the highest power of the variable is 2, forming a parabola or curve when graphed (e.g., y = x² + 3x - 5 or x² + y² = 25 for a circle).
Substitution method — replacing one variable in an equation with an expression from another equation to reduce two equations to one.
Discriminant — the value b² - 4ac from the quadratic formula, determining the number of real solutions (positive = two solutions, zero = one solution, negative = no real solutions).
Solution pair — a set of x and y values written as coordinates (x, y) that satisfies both equations simultaneously.
Rearranging — manipulating an equation to isolate one variable or to match a required form before substitution.
Core concepts
The structure of these problems
Simultaneous equations involving one linear and one quadratic always consist of:
- One straight line equation (e.g., y = 2x + 1 or 3x + y = 8)
- One curved equation (e.g., y = x² - 3x + 2 or x² + y² = 20)
The solutions represent the coordinates of intersection points between the line and the curve. Unlike linear simultaneous equations (which have one solution), these problems typically yield two solution pairs because a straight line usually intersects a parabola or circle at two distinct points.
The substitution method: step-by-step process
Identify which equation is easier to rearrange for y (or x). Usually, the linear equation is already in the form y = mx + c or can be rearranged easily.
Substitute the expression for y (or x) from the linear equation into the quadratic equation.
Simplify and expand to create a single quadratic equation in one variable.
Solve the quadratic equation using factorisation, the quadratic formula, or completing the square.
Substitute each solution back into the simpler linear equation to find corresponding values.
Write solution pairs as coordinates: (x₁, y₁) and (x₂, y₂).
When the linear equation needs rearranging
Not all linear equations arrive in the convenient form y = mx + c. You may encounter:
- Standard form: 2x + y = 5 → rearrange to y = 5 - 2x
- Both variables on one side: 3x - y = 4 → rearrange to y = 3x - 4
- x in terms of y: x = 2y + 1 → use this directly if the quadratic is in terms of x
Choose to express y in terms of x (or vice versa) based on which substitution creates the simplest quadratic to solve.
Types of quadratic equations encountered
After substitution and simplification, you'll need to solve various quadratic forms:
Factorisable quadratics: x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0
Difference of two squares: x² - 9 = 0 → (x - 3)(x + 3) = 0
Non-factorisable quadratics: Use the quadratic formula x = [-b ± √(b² - 4ac)] / 2a
Quadratics requiring completing the square: Particularly in non-calculator papers where the quadratic formula is less practical
Circle equations in simultaneous problems
AQA GCSE Mathematics includes problems where the quadratic is a circle equation: x² + y² = r²
The approach remains identical:
- Substitute the linear equation into the circle equation
- Expand brackets carefully (watch for (2x + 3)² = 4x² + 12x + 9, not 4x² + 9)
- Solve the resulting quadratic
- Find both coordinate pairs
Interpreting the number of solutions
The discriminant (b² - 4ac) reveals geometrically meaningful information:
- Two distinct solutions (b² - 4ac > 0): the line crosses the curve at two points
- One repeated solution (b² - 4ac = 0): the line is tangent to the curve (touches at exactly one point)
- No real solutions (b² - 4ac < 0): the line and curve don't intersect
Higher-tier AQA papers may ask you to prove a line is tangent to a curve by showing the discriminant equals zero.
Worked examples
Example 1: Standard simultaneous equations
Question: Solve the simultaneous equations:
y = x² - 4x + 1
y = 2x - 3
Solution:
Both equations are already expressed with y as the subject, so substitute the linear equation into the quadratic:
2x - 3 = x² - 4x + 1
Rearrange to standard quadratic form: 0 = x² - 4x + 1 - 2x + 3 0 = x² - 6x + 4
This doesn't factorise, so use the quadratic formula: x = [6 ± √(36 - 16)] / 2 x = [6 ± √20] / 2 x = [6 ± 2√5] / 2 x = 3 ± √5
x = 3 + √5 ≈ 5.236 or x = 3 - √5 ≈ 0.764
Substitute into y = 2x - 3:
When x = 3 + √5: y = 2(3 + √5) - 3 = 3 + 2√5 ≈ 7.472 When x = 3 - √5: y = 2(3 - √5) - 3 = 3 - 2√5 ≈ -1.472
Solution pairs: (3 + √5, 3 + 2√5) and (3 - √5, 3 - 2√5)
Or in decimal form: (5.24, 7.47) and (0.76, -1.47) [typically acceptable to 2 d.p. unless specified]
Example 2: Rearranging the linear equation first
Question: Solve the simultaneous equations:
x² + y² = 25
x + y = 7
Solution:
The second equation is linear. Rearrange for y: y = 7 - x
Substitute into the circle equation x² + y² = 25: x² + (7 - x)² = 25
Expand the bracket: x² + 49 - 14x + x² = 25
Simplify: 2x² - 14x + 49 = 25 2x² - 14x + 24 = 0
Divide by 2: x² - 7x + 12 = 0
Factorise: (x - 3)(x - 4) = 0
Therefore: x = 3 or x = 4
Substitute back into y = 7 - x:
When x = 3: y = 7 - 3 = 4 When x = 4: y = 7 - 4 = 3
Solution pairs: (3, 4) and (4, 3)
[These represent the two points where the line x + y = 7 crosses the circle with centre (0, 0) and radius 5]
Example 3: With fractions (Higher tier)
Question: Solve simultaneously:
y = x² + 2x - 5
2x + 3y = 6
Solution:
Rearrange the linear equation for y: 3y = 6 - 2x y = 2 - (2x/3)
Substitute into the quadratic: 2 - (2x/3) = x² + 2x - 5
Multiply through by 3 to eliminate fractions: 6 - 2x = 3x² + 6x - 15
Rearrange: 0 = 3x² + 6x + 2x - 15 - 6 0 = 3x² + 8x - 21
Use the quadratic formula: x = [-8 ± √(64 + 252)] / 6 x = [-8 ± √316] / 6 x = [-8 ± 2√79] / 6 x = [-4 ± √79] / 3
x ≈ 1.63 or x ≈ -4.30
Substitute into y = 2 - (2x/3):
When x = 1.63: y ≈ 2 - 1.09 ≈ 0.91 When x = -4.30: y ≈ 2 + 2.87 ≈ 4.87
Solution pairs: (1.63, 0.91) and (-4.30, 4.87) [to 2 d.p.]
Common mistakes and how to avoid them
• Forgetting to find both solutions — After solving for x (finding two values), students often substitute only one value back to find y. Always find y for each x value, creating two complete coordinate pairs. AQA mark schemes award separate marks for each coordinate pair.
• Sign errors when substituting expressions with brackets — When substituting y = 2x - 5 into x² + y² = 20, students write x² + 2x - 5² = 20 instead of x² + (2x - 5)² = 20. Always place substituted expressions in brackets before squaring or multiplying.
• Incorrect expansion of squared brackets — (3x + 2)² frequently becomes 9x² + 4 (missing the middle term). The correct expansion is 9x² + 12x + 4. Use the pattern (a + b)² = a² + 2ab + b² systematically.
• Solving for one variable only — Students solve the quadratic to find x = 2 and x = 5, then write these as final answers. The question asks to "solve the simultaneous equations", requiring complete coordinate pairs: (2, y₁) and (5, y₂). Read the command word carefully.
• Rearranging errors before substitution — From 2x + y = 8, writing y = 8 - x instead of y = 8 - 2x. Always perform the inverse operation on the x term correctly: subtract 2x from both sides gives y = 8 - 2x.
• Arithmetic errors in the quadratic formula — Particularly with the discriminant calculation or final simplification. In calculator papers, store intermediate values in calculator memory; in non-calculator papers, show discriminant calculation as a separate line to earn method marks even if the final answer is incorrect.
Exam technique for simultaneous equations involving one linear and one quadratic
• Command words: "Solve the simultaneous equations" or "Find the coordinates of the points of intersection" both require the same method. Questions worth 5-6 marks expect you to show full algebraic working through substitution, expansion, solving and back-substitution. Don't rely on graphical methods unless specifically asked.
• Mark allocation patterns: Typically 1 mark for correct substitution/rearrangement, 1-2 marks for forming and simplifying the quadratic correctly, 1 mark for solving the quadratic (both x values), 1 mark for each correct coordinate pair. Partial credit is available for correct method even with arithmetic slips, so show every step clearly.
• Calculator vs non-calculator papers: In non-calculator papers, expect quadratics that factorise neatly or have integer/simple surd solutions. Calculator paper questions may require the quadratic formula with decimal answers — give answers to 3 significant figures unless told otherwise, and use the "ANS" or memory function to avoid rounding errors in multi-step calculations.
• Space and layout: These questions require significant working space. If cramped, continue on additional space and indicate clearly. Write each major step on a new line: substitution line, expansion line, simplified quadratic, solution for x, substitution for y₁, substitution for y₂, final answer stated clearly.
Quick revision summary
To solve simultaneous equations with one linear and one quadratic: rearrange the linear equation to express y in terms of x (or vice versa), substitute this expression into the quadratic equation, expand and simplify to form a single quadratic, solve using factorisation or the quadratic formula to find two x values, substitute each x back into the linear equation to find corresponding y values, write solutions as two coordinate pairs. The solutions represent intersection points of the line and curve. Always show full working, find both complete pairs, and check solutions satisfy both original equations if time permits.