Question

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

$\frac{3x}{2}-\frac{5y}{3}+2=0$

$\frac{x}{3}+\frac{y}{2}=2\frac{1}{6}$

Answer 1

$\frac{x}{3}+\frac{y}{2}=2\frac{1}{6}\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\frac{(2x+3y)}{6}=\frac{13}{6}2x+3y=132x=13-3yx=\frac{(13-3y)}{2}\frac{3x}{2}-\frac{5y}{3}+2=0\frac{3x}{2}\left[\frac{(13-3y)}{2}\right]-\frac{5y}{3}=-2\frac{(39-9y)}{4}-\frac{5y}{3}=-2\frac{(117-27y-20y)}{12}=-2117-47y=-2447y=141y=3x=\frac{(13-3\times 3)}{2}x=\frac{4}{2}x=2$​​​​​​​