Simultaneous equations

Algebra

Questions marked Higher are aimed at the Higher tier.

These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.

Solve $x + y = 10$ and $x - y = 4$.

2 marks
Solve $x + y = 9$ and $x = 2y$.

2 marks
Solve $2x + y = 11$ and $x + y = 7$.

3 marks
Solve $3x + 2y = 16$ and $x + 2y = 8$.

3 marks
Solve $2x + 3y = 13$ and $4x - y = 5$.

4 marks
Two coffees and one tea cost £$5.50$. One coffee and two teas cost £$5$. Find the price of a coffee and the price of a tea.

4 marks
Solve $y = x^{2} - 4$ and $y = 3x$ simultaneously. Higher

5 marks
Solve $y = x + 2$ and $y = x^{2} - 4$ simultaneously. Higher

6 marks

Mark scheme

Answer. $x = 7$, $y = 3$
- Adding gives $2x = 14$.
- $x = 7$ and $y = 3$.
Answer. $x = 6$, $y = 3$
- $2y + y = 9$
- $y = 3$ and $x = 6$.
Answer. $x = 4$, $y = 3$
- Subtract to get $x = 4$.
- $y = 3$
Answer. $x = 4$, $y = 2$
- Subtract to get $2x = 8$.
- $x = 4$ and $y = 2$.
Answer. $x = 2$, $y = 3$
- Multiply the second equation by $3$ and add: $14x = 28$.
- $x = 2$ and $y = 3$.
Answer. Coffee £$2$, tea £$1.50$
- $2c + t = 5.50$ and $c + 2t = 5$.
- Doubling the second equation and subtracting gives $3t = 4.50$.
- Tea costs £$1.50$ and coffee costs £$2$.
Answer. $(4, 12)$ and $(-1, -3)$
- $x^{2} - 4 = 3x$
- $x^{2} - 3x - 4 = 0$, so $(x - 4)(x + 1) = 0$.
- The solutions are $(4, 12)$ and $(-1, -3)$.
Answer. $(3, 5)$ and $(-2, 0)$
- $x^{2} - 4 = x + 2$
- $x^{2} - x - 6 = 0$, so $(x - 3)(x + 2) = 0$.
- When $x = 3$, $y = 5$. When $x = -2$, $y = 0$.