How to solve simultaneous equations
When to eliminate and when to substitute, a worked example of each, and the sign error that costs more marks than anything else.
What the examiner is testing
Two equations, two unknowns, one pair of values that satisfies both at once. The marks are for a valid elimination or substitution step and for a correct pair — and there is usually a mark for checking, which most students skip.
Choosing a method
- Both equations look like \( ax + by = c \) → elimination.
- One equation already gives \( y = \ldots \) or \( x = \ldots \) → substitution.
- One equation is quadratic → substitution, always.
Worked example: elimination
$$ 5x + 2y = 16 $$
$$ 3x - 2y = 8 $$
Step 1 — match a coefficient. Here the \( y \) coefficients already match in size, at 2.
Step 2 — add or subtract. The signs are opposite, so add the equations:
$$ 8x = 24 \quad \Rightarrow \quad x = 3 $$
Signs the same, subtract. Signs opposite, add. Getting this backwards is the most common single error on the paper.
Step 3 — substitute back into the simpler original equation.
$$ 3(3) - 2y = 8 \quad \Rightarrow \quad 9 - 2y = 8 \quad \Rightarrow \quad y = 0.5 $$
Step 4 — check in the equation you did not use.
$$ 5(3) + 2(0.5) = 15 + 1 = 16 $$
It works, so the pair is right.
Answer: \( x = 3 \), \( y = 0.5 \).
When nothing matches
Multiply until something does. For \( 4x + 3y = 22 \) and \( 3x - 2y = 5 \), multiply the first by 2 and the second by 3 to make both \( y \) terms 6:
$$ 8x + 6y = 44 $$
$$ 9x - 6y = 15 $$
Every term gets multiplied, including the 22 and the 5. Now add as before.
Worked example: substitution
$$ y = 2x - 5 \qquad x^2 + y^2 = 25 $$
Substitute the first into the second:
$$ x^2 + (2x - 5)^2 = 25 $$
Expand carefully — \( (2x-5)^2 = 4x^2 - 20x + 25 \), not \( 4x^2 + 25 \):
$$ x^2 + 4x^2 - 20x + 25 = 25 \quad \Rightarrow \quad 5x^2 - 20x = 0 $$
There is no constant term, so factorise rather than reaching for the formula:
$$ 5x(x - 4) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 4 $$
Now find both \( y \) values from the linear equation: \( y = -5 \) and \( y = 3 \).
Answer: \( (0,\ -5) \) and \( (4,\ 3) \). Two points, because a line cuts a circle twice.
Check both in the circle: \( 0^2 + (-5)^2 = 25 \) and \( 4^2 + 3^2 = 25 \).
The three mistakes that lose marks
1. Adding when you should subtract. Same signs, subtract. Opposite signs, add. Write the sign of each coefficient down before you decide.
2. Multiplying only the left-hand side. Multiplying an equation by 3 multiplies the constant too.
3. Giving only one coordinate pair on a quadratic system. Two \( x \) values means two \( y \) values. Pair them by substituting into the linear equation, which is quicker and cannot introduce a false root.
30-second recap
Match a coefficient, add or subtract according to the signs, solve for one unknown, substitute back into the simpler equation, then check in the equation you have not used yet. If one equation is quadratic, substitute rather than eliminate, and expect two answer pairs.