Question

Solve the following pairs pf linear (simultaneous) equations using method of elimination by substitution:
$2x-3y+6=0$
$\Rightarrow 2x=3y-6$

Answer 1

$2x-3y+6=0$
$\Rightarrow 2x=3y-6$
$\Rightarrow x=\frac{3y-6}{2}.....\left(i\right)$
And,
$2x+3y-18=0$
$\Rightarrow 2x\left(\frac{3y-6}{2}\right)+3y=18..\left(from\right(i\left)\right)$
$\Rightarrow 3y-6+3y=18$
$\Rightarrow 6y=24$
$\Rightarrow y=4$
substituting the value of $y$ in (i) we have
$x=\frac{3\times 4-6}{2}=\frac{12-6}{2}=\frac{6}{2}=3$
$\therefore$ solution is $x=3$ and $y=4$