Simultaneous Equations

Why Simultaneous Equations?

Imagine you walk into a shop and buy 3 coffees and 2 teas for £11.50. Your friend buys 1 coffee and 4 teas for £10.00. Can you work out the price of each drink?

You have two unknowns and two pieces of information — that’s exactly when simultaneous equations come in. At GCSE you solved pairs of linear equations. At A-Level, the new challenge is solving a linear equation alongside a quadratic — and understanding what the solutions mean graphically.

Key Concepts

Linear-linear (GCSE recap)

You already know two methods for a pair of linear equations:

• Elimination — add or subtract the equations to remove one variable.
• Substitution — rearrange one equation and plug it into the other.

Both methods work for linear pairs. At A-Level, you’ll occasionally need these as a first step inside a bigger problem, so keep them sharp.

Linear-quadratic (the A-Level focus)

When one equation is quadratic (e.g. $x^2 + y^2 = 25$, or $y = x^2 - 3x + 1$), elimination won’t work. You must use substitution:

1. Rearrange the linear equation to make one variable the subject.
2. Substitute into the quadratic equation.
3. Solve the resulting quadratic (factorise, formula, or completing the square).
4. Substitute back into the linear equation to find the other variable.
5. Pair your answers: each $x$-value goes with its corresponding $y$-value.

Why substitution? Elimination relies on matching coefficients of a variable across two equations. When one equation has $x^2$ and the other has $x$, there is no way to eliminate — the powers don’t match.

The graphical picture

Solving simultaneous equations is the same as finding intersection points of two graphs. A line can intersect a quadratic curve in 0, 1, or 2 points:

• 2 solutions — the line crosses the curve twice.
• 1 solution — the line is tangent to the curve.
• 0 solutions — the line misses the curve entirely.

After substituting the linear equation into the quadratic, you get a single quadratic in one variable. The discriminant $b^2 - 4ac$ tells you which case you’re in:

$b^2 - 4ac > 0 \implies \text{2 solutions} \qquad b^2 - 4ac = 0 \implies \text{1 solution (tangent)} \qquad b^2 - 4ac < 0 \implies \text{no solutions}$

This connects directly to the quadratics and discriminant work you’ve already done.

Worked Examples

Example 1: Linear-linear by elimination (warm-up)

Solve simultaneously: $3x + 2y = 12$ and $x - 2y = 4$

Add the two equations to eliminate $y$:

$3x + 2y + x - 2y = 12 + 4$

$4x = 16 \implies x = 4$

Substitute into the second equation:

$4 - 2y = 4 \implies y = 0$

Solution: $(x, y) = (4, 0)$

Example 2: Linear + quadratic (the standard A-Level type)

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

Since both equations are written as $y = \ldots$, set them equal:

$x + 3 = x^2 + 1$

Rearrange to $0$:

$x^2 - x - 2 = 0$

Factorise:

$(x - 2)(x + 1) = 0$

$x = 2 \quad \text{or} \quad x = -1$

Substitute each $x$-value into the linear equation $y = x + 3$:

• When $x = 2$: $y = 2 + 3 = 5$
• When $x = -1$: $y = -1 + 3 = 2$

Solutions: $(x, y) = (2, 5)$ or $(-1, 2)$

Note: Always substitute back into the linear equation, not the quadratic. It’s quicker and you’re less likely to make errors.

Example 3: Linear + circle

Find the points of intersection of $x + y = 5$ and $x^2 + y^2 = 25$

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

Substitute into the circle equation:

$x^2 + (5 - x)^2 = 25$

$x^2 + 25 - 10x + x^2 = 25$

$2x^2 - 10x = 0$

$2x(x - 5) = 0$

$x = 0 \quad \text{or} \quad x = 5$

Substitute back into $y = 5 - x$:

• When $x = 0$: $y = 5$
• When $x = 5$: $y = 0$

Solutions: $(x, y) = (0, 5)$ or $(5, 0)$

Graphically, the line $x + y = 5$ meets the circle $x^2 + y^2 = 25$ at these two points on the axes.

Example 4: Using the discriminant to show no solutions

Show that the line $y = x + 6$ does not intersect the curve $y = x^2 + 3x + 8$.

Set the equations equal:

$x + 6 = x^2 + 3x + 8$

Rearrange:

$x^2 + 2x + 2 = 0$

Find the discriminant:

$b^2 - 4ac = (2)^2 - 4(1)(2) = 4 - 8 = -4$

Since $b^2 - 4ac < 0$, the quadratic has no real roots, so the line and curve do not intersect.

Example 5: Finding the value of $k$ for tangency

The line $y = 2x + k$ is tangent to the curve $y = x^2 + 4x + 1$. Find the value of $k$.

Set equal:

$2x + k = x^2 + 4x + 1$

Rearrange:

$x^2 + 2x + (1 - k) = 0$

For tangency (exactly one solution), set the discriminant equal to zero:

$b^2 - 4ac = 0$

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

$4 - 4 + 4k = 0$

$4k = 0 \implies k = 0$

So the tangent line is $y = 2x$. You can check: substituting gives $x^2 + 2x + 1 = 0$, i.e. $(x+1)^2 = 0$, which has exactly one solution $x = -1$, confirming tangency.

Common Mistakes to Avoid

“I’ll just eliminate the $x$” — You cannot use elimination when one equation is quadratic. The powers of $x$ (and $y$) don’t match across the two equations, so adding or subtracting them won’t cancel anything. You must substitute.

Forgetting to find the second variable — Solving the quadratic gives you values of one variable. You’re not finished. Substitute each value back into the linear equation to find the corresponding value of the other variable.

Unpaired answers — Writing $x = 2, -1$ and $y = 5, 2$ separately is ambiguous and will lose marks. Always present your answers as coordinate pairs: $(2, 5)$ and $(-1, 2)$.

Sign errors when rearranging — When you rearrange $y = 3 - 2x$ and substitute into $x^2 + y^2 = 10$, be careful expanding $(3 - 2x)^2$. Write out the bracket multiplication in full: $(3 - 2x)(3 - 2x) = 9 - 12x + 4x^2$.

Summary

The discriminant of the resulting quadratic tells you how many solutions to expect — and exam questions regularly ask you to use it to prove a line misses or is tangent to a curve.