Question

Find '$x+y$' for the following simultaneous equations : $2x+3y=12$ and $3x+2y=13$.

Answer 1

Let $2x+3y=12$ ...(1)
and $3x+2y=13$ ...(2)
From equation (2), find the value of $x$ in terms of $y$
$x=\frac{13-2y}{3}$ ...(3)

Substituting $x=\frac{13-2y}{3}$ in equation (1), we get

$\Rightarrow$ $2\left(\frac{13-2y}{3}\right)+3y=12$

$\Rightarrow$ $\frac{26-4y}{3}+3y=12$

$\Rightarrow \frac{26}{3}-\frac{4}{3}y+3y=12$

$\Rightarrow 3y-\frac{4y}{3}=12-\frac{26}{3}$

$\Rightarrow \frac{5y}{3}=\frac{10}{3}$

$\Rightarrow y=\frac{10}{5}=2$
$\Rightarrow y=2$
Substituting $y=2$ in equation (3), we get

$x=\frac{13-2\times 2}{3}$

$\Rightarrow x=\frac{13-4}{3}$

$\Rightarrow x=\frac{9}{3}=3$

$\Rightarrow x=3$
Therefore, the value of $x+y=3+2$ is $5$.