Number the equations.

$\begin{array}{rcl} 5 a + 2 c & = & 10 (1) \\ 2 a - 4 c & = & 7 (2) \end{array}$

Make the c terms equal by multiplying all parts of equation (1) by 2 and leaving the second equation as it is.
This will give two 4c terms with opposite signs.

$\begin{array}{rcl} 10 a + 4 c & = & 20 (3) \end{array}$

[1]

Eliminate the x terms by adding equations (2) and (3).
Be careful to add every term and that adding -4c is the same as subtracting 4c.

$12 a = 27$

Solve the equation to find a by dividing both sides by 12.

$\begin{array}{rcl} a & = & \frac{27}{12} = \frac{9}{4} = 2.25 \end{array}$

Substitute $a = 2.25$ into either of the two original equations.

$(1) 5 (2.25) + 2 c = 10$

Solve this equation to find c.

$\begin{array}{rcl} 11.25 + 2 c & = & 10 \\ 2 c & = & - 1.25 \\ c & = & - 0.625 \end{array}$

[1]

Substitute $a = 2.25$ and $c = - 0.625$ into the other original equation to check that they are correct.

$\begin{array}{rcl} (2) 2 a - 4 c & = & 2 (2.25) - 4 (- 0.625) \\ = & 4.5 + 2.5 \\ = & 7 \end{array}$

Be sure to get a and c the right way round for the final answer!

$a = 2.25, c = - 0.625$ [1]

Pattern number	1	2	3	4
Number of grey mats	6	9	12	15
Total number of mats	6	10	15	21

Simultaneous Equations (Cambridge O Level Maths)

Exam Questions