Question

Solve for $x$ and $y$:
$\frac{5}{x}+6y=13\frac{3}{x}+4y=7(x\ne 0)$

Answer 1

Given

$\frac{5}{x}+6y=13...\left(1\right)\frac{3}{x}+4y=7...\left(2\right)$

Let us use elimination method to solve the given system of equation.

Multiply equation $\left(1\right)$ by $2$ and equation $\left(2\right)$ by $3$, we get:

$\frac{10}{x}+12y=26...\left(3\right)\frac{9}{x}+12y=21...\left(4\right)$

Now subtracting equation $\left(4\right)$ from $\left(3\right)$

$(\frac{10}{x}+12y)-(\frac{9}{x}+12y)=26-21$

$\Rightarrow \frac{1}{x}=5$

$\Rightarrow x=\frac{1}{5}$

Substituting the value of $x$ in equation $\left(1\right)$, we get:

$5\left(\frac{1}{5}\right)+6y=13$

$\Rightarrow y=-2$

Hence, $x=\frac{1}{5}$ and $y=-2$.