Question

Given:
$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$
$\frac{3}{x}+\frac{2}{y}=0$
Find 'a' for which $y=ax-4$

• A. 1
• B. 2
• C. 3
• D. None of the above

Answer 1

We have, $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}\dots \dots \left(1\right)$

$\frac{3}{x}+\frac{2}{y}=0\dots \dots \left(2\right)$

Putting $\frac{1}{x}=u$ and $\frac{1}{v}=u$, we get

$2u+\frac{2}{3}v=\frac{1}{6}\dots \dots \left(3\right)$

$3u+2v=0\dots \dots \left(4\right)$

Multiplying equation (4) by $\frac{1}{3}$, we get

$u+\frac{2}{3}v=0\dots \dots \left(5\right)$

Subtracting equation (5) from equation (3), we get

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

Putting $u=\frac{1}{6}$ in (4), we get

$3\left(\frac{1}{6}\right)+2v-0\Rightarrow \frac{1}{2}+2v-0$

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

Now, $u=\frac{1}{6}\Rightarrow \frac{1}{x}=\frac{1}{6}\Rightarrow x=6$

and, $v=-\frac{1}{4}\Rightarrow \frac{1}{y}=-\frac{1}{4}\Rightarrow y=-4$

Hence, the solution is x = 6, y = -4. Again,

$y=ax-4\Rightarrow -4\Rightarrow a\left(6\right)-4$

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

$\Rightarrow a=\frac{0}{6}=0$