Question

Solve: $3(2x+y)=7xy$; $3(x+3y)=11xy$ using elimination method.

Answer 1

Given:$3(2x+y)=7xy,3(x+3y)=11xy$

Divide both sides by $xy$

$\frac{6x+3y}{xy}=7,\frac{3x+9y}{xy}=11$

$\frac{6}{y}+\frac{3}{x}=7,\frac{3}{y}+\frac{9}{x}=11$

Let $u=\frac{1}{x},v=\frac{1}{y}$ then the above equations becomes

$6v+3u=7$ ......$\left(1\right)$ and $3v+9u=11$ ....$\left(2\right)$

Eqn$\left(1\right)-2\times \left(2\right)$

$\Rightarrow 6v+3u-6v-18u=7-22$

$\Rightarrow -15u=-15$ or $u=1$

Substitute the value of $u=1$ in eqn$\left(1\right)$ we get

$6v+3u=7$ or $6v=7-3=4$ or $v=\frac{2}{3}$

From above we have $u=\frac{1}{x}\Rightarrow x=\frac{1}{u}=1$

and $v=\frac{1}{y}\Rightarrow y=\frac{1}{v}=\frac{3}{2}$

$\therefore x=1,y=\frac{3}{2}$