Question

Solve the following systems of equations by using the method of substitution,
$2x+3y=9$ and $3x+4y=5$.

Answer 1

The given system of equations is
$2x+3y=9$
$3x+4y=5$

From equation (i), we get

$3y=9-2x\Rightarrow y=\frac{9-2x}{3}$

Substituting $y=\frac{9-2x}{3}$ in equation (ii), we get

$3x+4\left(\frac{9-2x}{3}\right)=5$

$\Rightarrow \frac{9x+36-8x}{3}=5$

$\Rightarrow x+36=15$

$\Rightarrow x=-21$

Putting $x=-21$ in $y=\frac{9-2x}{3}$, we get

$y=\frac{9+42}{3}=17$

Hence, the solution of the given system of equations is $x=-21,y=17$.