Simultaneous Equations: One Linear, One Quadratic

Introduction

When two equations must be satisfied simultaneously and one is linear while the other is quadratic, the solution gives the coordinates of the point(s) where a straight line meets a curve (such as a circle or parabola). This is a guaranteed topic in 9709 Paper 1 — it appears either as a standalone question or embedded within geometry questions about circles and intersection. The examiners' preferred method is substitution, and that is the only method required by the syllabus.

Core Concept

A linear equation contains no term of degree higher than 1 (e.g. $x + y + 1 = 0$ ). A quadratic equation contains at least one term of degree 2 (e.g. $x^2 + y^2 = 25$ ). Solving them simultaneously means finding all pairs $(x, y)$ that satisfy both equations at once.

Why substitution? Elimination works neatly for two linear equations, but once a quadratic is involved, the algebra demands that you express one variable in terms of the other from the linear equation and substitute into the quadratic. This converts the problem into a single-variable quadratic, which you already know how to solve.

Step-by-step procedure:

Rearrange the linear equation to make one variable the subject (choose whichever is simpler — avoid fractions where possible).
Substitute the expression into the quadratic equation.
Expand and simplify to obtain a quadratic equation in one variable.
Solve the resulting quadratic (by factorisation, completing the square, or the quadratic formula).
Back-substitute each solution into the linear equation to find the corresponding value of the other variable.
State the solution pairs clearly, pairing each $x$ -value with its matching $y$ -value.

The number of solutions corresponds to the number of intersection points:

Discriminant $\Delta = b^2 - 4ac$	Geometric interpretation
$\Delta > 0$	Two distinct intersection points
$\Delta = 0$	Line is tangent to curve (one repeated point)
$\Delta < 0$	No real intersections

Key Formulae & Definitions

Quadratic formula (used when factorisation is not straightforward):

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

for a quadratic $ax^2 + bx + c = 0$ .

Discriminant:

\Delta = b^2 - 4ac

General substitution setup: Given the linear equation $y = mx + c$ and the quadratic $f(x, y) = 0$ , substitute $y = mx + c$ to obtain $f(x,\, mx+c) = 0$ .

Worked Examples

Example 1 — Classic circle and line (syllabus example)

Solve simultaneously: $x + y + 1 = 0$ and $x^2 + y^2 = 25$ .

Step 1 — Rearrange the linear equation for $y$ :

y = -x - 1

Step 2 — Substitute into the quadratic:

x^2 + (-x-1)^2 = 25

Step 3 — Expand:

x^2 + (x^2 + 2x + 1) = 25

2x^2 + 2x + 1 = 25

Step 4 — Rearrange to standard form:

2x^2 + 2x - 24 = 0

x^2 + x - 12 = 0

Step 5 — Factorise:

(x + 4)(x - 3) = 0

x = -4 \quad \text{or} \quad x = 3

Step 6 — Back-substitute using $y = -x - 1$ :

When $x = -4$ : $\quad y = -(-4) - 1 = 3$
When $x = 3$ : $\quad y = -(3) - 1 = -4$

Solution: $(x, y) = (-4,\; 3)$ and $(x, y) = (3,\; -4)$ .

Example 2 — Parabola and line

Solve simultaneously: $y = 2x + 1$ and $y = x^2 - x - 3$ .

Step 1 — The linear equation already gives $y$ explicitly:

y = 2x + 1

Step 2 — Substitute into the quadratic:

2x + 1 = x^2 - x - 3

Step 3 — Rearrange to standard form:

0 = x^2 - x - 3 - 2x - 1

x^2 - 3x - 4 = 0

Step 4 — Factorise:

(x - 4)(x + 1) = 0

x = 4 \quad \text{or} \quad x = -1

Step 5 — Back-substitute using $y = 2x + 1$ :

When $x = 4$ : $\quad y = 2(4) + 1 = 9$
When $x = -1$ : $\quad y = 2(-1) + 1 = -1$

Solution: $(x, y) = (4,\; 9)$ and $(x, y) = (-1,\; -1)$ .

Example 3 — Tangency condition

Show that the line $y = 3x - 5$ is a tangent to the curve $y = x^2 - 2x$ , and find the point of tangency.

Step 1 — Substitute $y = 3x - 5$ into $y = x^2 - 2x$ :

3x - 5 = x^2 - 2x

Step 2 — Rearrange:

x^2 - 5x + 5 = 0 \longrightarrow \text{wait — rearranging correctly:}

x^2 - 2x - 3x + 5 = 0 \implies x^2 - 5x + 5 = 0

Step 3 — Compute the discriminant:

\Delta = (-5)^2 - 4(1)(5) = 25 - 20 = 5

This is not zero, so let us reconsider the problem with $y = 3x - \tfrac{9}{4}$ — actually, let us keep the integrity of the algebra. The line $y = 3x - 5$ meets the curve at two points (since $\Delta = 5 > 0$ ).

Revised Example 3 — Tangency:

Show that $y = 2x - 1$ is a tangent to $y = x^2 - 2x + 3$ , and find the point of tangency.

Step 1 — Substitute:

2x - 1 = x^2 - 2x + 3

Step 2 — Rearrange:

x^2 - 4x + 4 = 0

Step 3 — Discriminant:

\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0

Since $\Delta = 0$ , the line is indeed a tangent.

Step 4 — Solve: $(x-2)^2 = 0 \implies x = 2$

Step 5 — Back-substitute: $y = 2(2) - 1 = 3$

Point of tangency: $(2,\; 3)$ .

Common Mistakes & Examiner Pitfalls

Substituting into the wrong equation. Always rearrange the linear equation and substitute into the quadratic. Substituting the other way leads to a more complex (or impossible) rearrangement.
Incomplete expansion of brackets. When substituting $y = -x - 1$ into $y^2$ , students often write $(-x-1)^2 = x^2 + 1$ (forgetting the cross term $2x$ ). Expand carefully: $(-x-1)^2 = x^2 + 2x + 1$ .
Mismatching solution pairs. After finding two $x$ -values, always substitute back into the linear equation (not the quadratic) to find each $y$ . Mixing up pairs is a common 1-mark error.
Missing the second solution. A quadratic typically yields two solutions. Do not stop after finding one.
Arithmetic sign errors when rearranging. When moving terms across the equals sign, take care with negative signs — especially in expressions like $x^2 - (mx + c)^2$ .
Not checking the discriminant for tangency proofs. If asked to "show that a line is a tangent," you must explicitly state $\Delta = 0$ and conclude from it.

Practice Questions

Q1. Solve simultaneously: $y = x - 3$ and $x^2 + y^2 = 29$ .

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Substitute $y = x - 3$ into $x^2 + y^2 = 29$ :

x^2 + (x-3)^2 = 29

x^2 + x^2 - 6x + 9 = 29

2x^2 - 6x - 20 = 0

x^2 - 3x - 10 = 0

(x-5)(x+2) = 0

$x = 5$ or $x = -2$ .

$x = 5$ : $y = 5 - 3 = 2$ $\implies (5, 2)$
$x = -2$ : $y = -2 - 3 = -5$ $\implies (-2, -5)$

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Q2. Find the values of $x$ and $y$ satisfying $2x - y = 3$ and $y = x^2 - 4x + 1$ .

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Rearrange the linear equation: $y = 2x - 3$ .

Substitute into the quadratic:

2x - 3 = x^2 - 4x + 1

x^2 - 6x + 4 = 0

Using the quadratic formula:

x = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}

$x = 3 + \sqrt{5}$ : $y = 2(3+\sqrt{5}) - 3 = 3 + 2\sqrt{5}$
$x = 3 - \sqrt{5}$ : $y = 2(3-\sqrt{5}) - 3 = 3 - 2\sqrt{5}$

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Q3. Determine the number of intersections between the line $x + 2y = 5$ and the curve $x^2 + 3y^2 = 12$ without fully solving.

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Rearrange the linear equation: $x = 5 - 2y$ .

Substitute:

(5-2y)^2 + 3y^2 = 12

25 - 20y + 4y^2 + 3y^2 = 12

7y^2 - 20y + 13 = 0

Discriminant: $\Delta = (-20)^2 - 4(7)(13) = 400 - 364 = 36 > 0$ .

Since $\Delta > 0$ , there are two distinct intersection points.

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Q4. The line $y = kx - 2$ is a tangent to the curve $y = x^2 + 4x + 2$ . Find the possible values of $k$ .

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Substitute $y = kx - 2$ into $y = x^2 + 4x + 2$ :

kx - 2 = x^2 + 4x + 2

x^2 + (4-k)x + 4 = 0

For tangency, $\Delta = 0$ :

(4-k)^2 - 4(1)(4) = 0

16 - 8k + k^2 - 16 = 0

k^2 - 8k = 0

k(k - 8) = 0

$k = 0$ or $k = 8$ .

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Q5. Solve simultaneously: $3x + y = 7$ and $x^2 + xy = 6$ .

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Rearrange the linear equation: $y = 7 - 3x$ .

Substitute into $x^2 + xy = 6$ :

x^2 + x(7 - 3x) = 6

x^2 + 7x - 3x^2 = 6

-2x^2 + 7x - 6 = 0

2x^2 - 7x + 6 = 0

(2x - 3)(x - 2) = 0

$x = \tfrac{3}{2}$ or $x = 2$ .

$x = \tfrac{3}{2}$ : $y = 7 - 3 \cdot \tfrac{3}{2} = 7 - \tfrac{9}{2} = \tfrac{5}{2}$ $\implies \left(\tfrac{3}{2},\, \tfrac{5}{2}\right)$
$x = 2$ : $y = 7 - 6 = 1$ $\implies (2,\, 1)$

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Connections

Prerequisite topics (assumed known):

Solving Quadratic Equations — factorisation, completing the square, and the quadratic formula are essential once substitution reduces the system to a single quadratic.
Rearranging Algebraic Expressions — fluency with making a variable the subject is required in Step 1.

Likely next subtopics:

Discriminant and Nature of Roots — the tangency condition ( $\Delta = 0$ ) in these problems is a direct application of the discriminant; this connection deepens your understanding of both topics.
Coordinate Geometry: Circles — many 9709 circle questions require solving a line–circle system, which is precisely the $x + y + 1 = 0$ with $x^2 + y^2 = 25$ structure practised here.
Quadratic Inequalities — once you can solve these systems, inequality questions ask for the range of $x$ or $y$ satisfying both constraints, extending the same substitution technique.
Functions and Graphs — understanding intersections of curves as solutions to simultaneous equations reinforces the graphical interpretation of all equation-solving in Pure Mathematics.