Question

Solve :

$\frac{2}{x}+\frac{2}{3y}$ = $\frac{1}{6}$
$\frac{3}{x}+\frac{2}{y}$ = 0
And hence find $a$ for which $y=ax-4$

Answer 1

Given that

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

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

On substitute $\frac{1}{x}=u,\frac{1}{y}=v,$

we have

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

$12u+4v=1$ $($ multiply both side by $6$$)$ -------------$\left(1\right)$

Also$,$ $3u+2v=0$-------------------$\left(2\right)$

Apply $e{q}^n\left(1\right)-e{q}^n\left(2\right)\times 4$

$\left(12u+4v\right)-\left(12u+8v\right)=1-0$

$-4v=1$

$v=\frac{-1}{4}$

$u=\frac{-2v}{3}$ $($using $\left(2\right)$ $)$

$u=\frac{-2}{3}\times \left(\frac{-1}{4}\right)=\frac{1}{6}$

$\therefore x=\frac{1}{u}=6$

$y=\frac{1}{v}=-4$

Since,

$y=ax-4$

$-4=a\times 6-4$

$a=0$

Hence, this is the answer.