Question

Solve the following simultaneous equations.
$3a+5b=26;a+5b=22$

Answer 1

$3a+5b=\mathrm{26.......}\left(1\right)a+5b=\mathrm{22.......}\left(2\right)$

Subtracting equation (2) from (1), we get

$\Rightarrow 3a+5b-(a+5b)=26-22$

$\Rightarrow 3a+5b-a-5b=4$

$\Rightarrow 2a=4$

$\therefore a=2$

Now, $3\times 2+5b=26$

$\Rightarrow 5b=20$

$\Rightarrow b=4$

Hence, $a=2$ and $b=4$