Solving Simultaneous Equations Algebraically and Graphically Revision Notes | GCSE AQA Higher Further Maths

1. Why Simultaneous Equations Matter

One equation with two unknowns has infinitely many solutions. Add a second equation with the same two unknowns, and suddenly there is (usually) exactly one pair of values that satisfies both at once. That is the power of simultaneous equations.

You will meet two algebraic methods — elimination and substitution — and a graphical approach. The specification requires you to handle two linear equations and also the harder case where one equation is linear and one is quadratic (second order). Both cases appear at GCSE.

2. Solving Two Linear Equations — Elimination

Elimination works by making the coefficient of one variable identical in both equations, then adding or subtracting to cancel it out. Think of it as a balancing act: whatever you do to one side, you do to the other, so the equality is preserved.

Worked example: Elimination: Two Linear Equations

Solve simultaneously: $3x + 2y = 16$ and $5x - 2y = 8$

1. Equation (1): $3x + 2y = 16$ Equation (2): $5x - 2y = 8$ — Labelling keeps working organised and earns method marks [M1].
2. Both equations have $2y$ — one positive, one negative. — When signs are opposite, add the equations to eliminate $y$.
3. $(3x + 2y) + (5x - 2y) = 16 + 8$ $8x = 24$ — $+2y$ and $-2y$ cancel. [M1] for correct elimination step.
4. $x = \frac{24}{8} = 3$ — [A1] for $x = 3$.
5. Substitute $x = 3$ into equation (1): $3(3) + 2y = 16$ $9 + 2y = 16$ $2y = 7$ $y = 3.5$ — Always substitute into the simpler original equation. [A1] for $y = 3.5$.
6. $5(3) - 2(3.5) = 15 - 7 = 8$ ✓ — Checking in the unused equation confirms both values are correct. This step takes seconds and saves marks.

Answer: $x = 3$, $y = 3.5$

When the coefficients do not match, multiply one or both equations first. For example, if one equation has $3y$ and the other has $2y$, multiply the first by 2 and the second by 3 to produce $6y$ in both — then subtract.

3. Solving Two Linear Equations — Substitution

Substitution is particularly useful when one equation is already solved for a variable — for instance, $y = 2x - 1$. You feed that expression directly into the other equation, replacing $y$ everywhere it appears.

Worked example: Substitution: Two Linear Equations

Solve simultaneously: $y = 2x - 1$ and $3x + 4y = 18$

1. Equation (1): $y = 2x - 1$ Equation (2): $3x + 4y = 18$ — Equation (1) is already in the form $y =...$, so substitution is the natural choice.
2. $3x + 4(2x - 1) = 18$ — Replace every $y$ in equation (2) with $(2x - 1)$. [M1] for correct substitution.
3. $3x + 8x - 4 = 18$ $11x - 4 = 18$ $11x = 22$ — Expand the bracket carefully — a sign error here is the most common mistake.
4. $x = \frac{22}{11} = 2$ — [A1] for $x = 2$.
5. $y = 2(2) - 1 = 4 - 1 = 3$ — [A1] for $y = 3$.
6. $3(2) + 4(3) = 6 + 12 = 18$ ✓ — Both values satisfy the original equation — solution confirmed.

Answer: $x = 2$, $y = 3$

4. One Linear, One Quadratic — Substitution is Essential

When one equation is quadratic (second order), elimination rarely works cleanly. Substitution is the standard approach: rearrange the linear equation for one variable, substitute into the quadratic, and solve the resulting quadratic equation.

A linear-quadratic pair can have two solutions, one solution (where the line is a tangent to the curve), or no real solutions. Expect two solutions in most GCSE questions, but be ready to recognise the other cases.

Worked example: Linear and Quadratic: Two Solutions

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

1. Equation (1): $y = x + 3$ (linear) Equation (2): $y = x^2 - 2x - 1$ (quadratic) — The linear equation is already solved for $y$, so substitution is immediate.
2. $x + 3 = x^2 - 2x - 1$ — Replace $y$ in equation (2) with $x + 3$. [M1] for correct substitution.
3. $0 = x^2 - 2x - 1 - x - 3$ $0 = x^2 - 3x - 4$ — Collect all terms on one side. Bring everything to the right so the quadratic equals zero.
4. $x^2 - 3x - 4 = (x - 4)(x + 1)$ — Check: $-4 \times 1 = -4$ and $-4 + 1 = -3$. [M1] for correct factorisation.
5. $x - 4 = 0 \Rightarrow x = 4$ $x + 1 = 0 \Rightarrow x = -1$ — [A1] for both values of $x$.
6. When $x = 4$: $y = 4 + 3 = 7$ When $x = -1$: $y = -1 + 3 = 2$ — Always use the linear equation to find $y$ — it is simpler and less error-prone. [A1] for both pairs.
7. $x = 4$: $4^2 - 2(4) - 1 = 16 - 8 - 1 = 7$ ✓ $x = -1$: $(-1)^2 - 2(-1) - 1 = 1 + 2 - 1 = 2$ ✓ — Both pairs satisfy the quadratic. Note $(-1)^2 = (-1) \times (-1) = 1$, not $-1$.

Answer: $(4, 7)$ and $(-1, 2)$

The graph above shows the line $y = x + 3$ crossing the parabola $y = x^2 - 2x - 1$ at exactly two points. Those intersection points are $(-1, 2)$ and $(4, 7)$ — the same answers the algebra gave. This is the graphical interpretation of the solution.

5. Graphical Solutions and What They Mean

Every solution to a pair of simultaneous equations corresponds to a point where the two graphs cross. For two straight lines, there is normally one intersection point — one solution. For a line and a parabola, there can be zero, one, or two intersection points.

On an exam question asking you to solve graphically, read off the $x$ and $y$ coordinates of each intersection point and state them as a pair. If the question says 'solve algebraically', do not rely on a sketch — show full algebraic working.

6. Practice Problems

The methods are clear — now it is time to apply them. Work through each problem before checking the solution. Pay particular attention to sign handling when rearranging into quadratic form.

7. Exam Technique and Common Mistakes

Most marks lost on simultaneous equation questions come from a small set of predictable errors. Knowing them in advance means you can actively guard against them under exam conditions.

Simultaneous equations — whether both linear or one linear and one quadratic — always reduce to the same core skill: use one equation to eliminate a variable from the other, then solve what remains. The graphical picture (intersection points) and the algebraic answer must always agree, and checking one against the other is a reliable way to catch errors before the examiner does.