Question

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

$\frac{x}{6}+\frac{y}{15}=4$

$\frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$

Answer 1

$\frac{x}{6}+\frac{y}{15}=4\frac{(5x+2y)}{30}=45x+2y=1205x=120-2yx=\frac{(120-2y)}{5}\frac{x}{3}-\frac{y}{12}=4\frac{3}{4}\frac{1}{3}[x-\frac{y}{4}]=\frac{19}{4}\frac{1}{3}[\frac{(120-2y)}{5}-\frac{y}{4}]=\frac{19}{4}\frac{(480-13y)}{20}=\frac{57}{4}480-13y=28513y=195y=15x=\frac{(120-2\times 15)}{5}x=\frac{90}{5}x=18$