Question

Solve the following system of equations by the method of substitution:
$2p+3q=-5$, $3p-2q=12$

Answer 1

Given equations are

$2p+3q=-\mathrm{5......}\left(1\right)$

and $3p-2q=\mathrm{12......}\left(2\right)$

Therefore, $3p=12+2q\Rightarrow p=\frac{12+2q}{3}.......\left(3\right)$

Substitute this value of $p$ in equation $\left(1\right)$, we get

$2\left(\frac{12+2q}{3}\right)+3q=-5$

$\Rightarrow 24+4q+9q=-15$

$\Rightarrow 13q=-39$

$\Rightarrow q=-3$

Put this value of $q$ in equation $\left(3\right)$.

$p=\frac{12+2(-3)}{3}$

$\Rightarrow p=\frac{12-6}{3}$

$\Rightarrow p=2$

Therefore, the solution is $p=2,q=-3$.