A-Level — mathsStandard2 min read

Simultaneous Equations

Solve simultaneous equations at A-Level: linear, quadratic, and parametric systems. Interpret solutions graphically.

Tutor AI Team2026-02-25T01:53:03.156Z

A-Level simultaneous equations extend GCSE by including quadratic and more complex systems. You must solve algebraically and interpret graphically.

Methods

Linear Systems

Use elimination or substitution. Same as GCSE but with more complex expressions.

Linear-Quadratic Systems

Substitute the linear equation into the quadratic. Always use substitution.

$y = 2x + 1$ and $x^2 + y^2 = 10$

$x^2 + (2x+1)^2 = 10$ → $5x^2 + 4x - 9 = 0$ → $(5x+9)(x-1)=0$

$x = 1, y = 3$ or $x = -\frac{9}{5}, y = -\frac{13}{5}$

Graphical Interpretation

Two solutions: line crosses curve twice.
One solution (tangent): discriminant = 0.
No solutions: line doesn't meet curve.

The Discriminant Test

After substitution, if the resulting quadratic $ax^2 + bx + c = 0$ has:

$b^2 - 4ac > 0$ : two intersections
$b^2 - 4ac = 0$ : tangent
$b^2 - 4ac < 0$ : no intersection

Worked Example: Example 1

Problem

Find where $y = x + 3$ meets $y = x^2 + 1$ .

$x + 3 = x^2 + 1$ → $x^2 - x - 2 = 0$ → $(x-2)(x+1) = 0$

$x = 2, y = 5$ and $x = -1, y = 2$ .

Solution

Worked Example: Example 2

Problem

Show $y = 3x + k$ is tangent to $y = x^2$ when $k = -\frac{9}{4}$ .

$3x + k = x^2$ → $x^2 - 3x - k = 0$ . Discriminant: $9 + 4k = 0$ → $k = -\frac{9}{4}$ .

Solution

Practice Problems

1. Solve $y = x - 1$ and $x^2 + y^2 = 25$ .
1. Find values of $k$ for which $y = 2x + k$ is tangent to $y = x^2 + 3x$ .

Want to check your answers and get step-by-step solutions?

Key Takeaways

✓
Substitution for linear-quadratic systems.
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Discriminant determines number of intersections.
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$\Delta = 0$ means the line is a tangent to the curve.

Tags:

#a-level#maths#pure#algebra#simultaneous-equations

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