Question

Solve $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$ and $\frac{3}{x}+\frac{2}{y}=0$ and hence find 'a' for which, $y=ax-4$.

Answer 1

As, $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$ and $\frac{3}{x}+\frac{2}{y}=0$ are not in the general form of linear equations. To make them into linear equaions,

Take $\frac{1}{x}=u$ and $\frac{1}{y}=v$. The given system of equations become

$2u+\frac{2}{3}v=\frac{1}{6}$

$\Rightarrow 12u+4v=1$ (i)
and, $3u+2v=0$ .(ii)

Multiplying equation (ii) by $2$ and subtracting from equation (i), we get
$6u-1\Rightarrow u=\frac{1}{6}$

Putting $u=\frac{1}{6}$ in (i), we get

$2+4v=1\Rightarrow v=-\frac{1}{4}$

Hence, $x=\frac{1}{u}=6$ and $y=\frac{1}{v}=-4$

So, the solution of the given system of equations is $x=6,y=-4$

Putting $x=6,y=-4$ in $y=ax-4$, we get

$-4=6a-4\Rightarrow a=0$.