Question

Solve for $x$ and $y$:
$\frac{44}{x+y}+\frac{30}{x-y}=10$
$\frac{55}{x+y}+\frac{40}{x-y}=13$
where $x+y\ne 0$ and $x-y\ne 0$.

Answer 1

$\frac{44}{x+y}+\frac{30}{x-y}=10$

$\frac{55}{x+y}+\frac{40}{x-y}=13$

Let $\frac{1}{x+y}=m;\frac{1}{x-y}=n$

$\therefore 44m+30n=10$ ……$\left(1\right)$

$55m+40n=13$ ……$\left(2\right)$

Now multiplying $\left(1\right)$ by $4$ and $\left(2\right)$ by $3$

$\therefore 176m+120n=40$

$-165m+120n=39$

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$11m$ $=1$

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$\therefore m=\frac{1}{11}$

$\therefore n=\frac{1}{5}$

$\therefore x+y=11$

$x-y=5$

$\therefore 2x=16$

$\therefore x=8$

$\therefore y=3$.