Question

Solve the following simultaneous equations:
$\frac{10}{x+y}+\frac{2}{x-y}=4$;
$\frac{5}{x+y}-\frac{5}{3(x-y)}=\frac{-2}{3}$

Answer 1

Given, $\frac{10}{x+y}+\frac{2}{x-y}=4$

$\frac{5}{x+y}-\frac{5}{3(x-y)}=-\frac{2}{3}$

Substituting $\frac{1}{x+y}=a$ and $\frac{1}{x-y}=b$ in the above equations

$\therefore$ $10a+2b=\mathrm{4........}\left(1\right)$
$15a-5b=-\mathrm{2..........}\left(2\right)$

Multiplying equations $\left(1\right)$ by $5$ and $\left(2\right)$ by $2$
$5(10a+2b=4)$
$50a+10b=\mathrm{20........}\left(3\right)$
$2(15a-5b=-2)$
$30a-10b=-\mathrm{4..........}\left(4\right)$

Adding equation $\left(3\right)$ and $\left(4\right)$
$50a+10b=20$
$30a-10b=-4$
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$80a=16$
$a=\frac{16}{80}$
$\therefore$ $a=\frac{1}{5}$
Substituting $a=\frac{1}{5}$ on equation $\left(1\right)$
$10a+2b=4$
$10\times \frac{1}{5}+2b=4$
$2+2b=4$
$2b=4-2$
$2b=2$
$b=1$

$\frac{1}{x+y}=\frac{1}{5}$, $\frac{1}{x-y}=1$

$x+y=\mathrm{5.........}\left(5\right)$
$x-y=\mathrm{1..........}\left(6\right)$
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$2x=6$ .......... (Adding)
$x=\frac{6}{2}$
$x=3$

Substituting $x=3$ in equation $\left(5\right)$ we get
$x+y=5$
$3+y=5$
$y=5-3$
$\therefore$ $y=2$
Solution set: $x=3;y=2$