Question

Solve the following systems of equations:

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

$\frac{15}{x+y}-\frac{9}{x-y}=-2$

Answer 1

Let $\frac{1}{x+y}$=t and $\frac{1}{x-y}$=s

Substitute it in given equations, we get

10t+2s=4 ... (1)

15t-9s=-2 ... (2)

Multiplying equation (1) with 3 and equation (2) with 2

$\Rightarrow$ 30t +6s =12 ... (3)

$\Rightarrow$ 30t -18s =-4 ... (4)

(4) - (3)

$\Rightarrow$ 30t-18s-30t-6s=-4-12

$\Rightarrow$ -24s =-16

$\Rightarrow$ s =$\frac{-16}{-24}$

$\Rightarrow$ s= $\frac{2}{3}$

Substitute s =$\frac{2}{3}$ in equation (2)

$\Rightarrow$ 15t -9($\frac{2}{3}$)=-2

$\Rightarrow$ 15t -6=-2

$\Rightarrow$ 15t = -2+6

$\Rightarrow$ 15t =4

$\Rightarrow$ t = $\frac{4}{15}$

We know that

$\Rightarrow$ $\frac{1}{x+y}$=t

$\Rightarrow$ $\frac{1}{x+y}$= $\frac{4}{15}$

$\Rightarrow$ x + y = $\frac{15}{4}$ ... (5)

Also we know that

$\Rightarrow$ $\frac{1}{x-y}$=s

$\Rightarrow$ $\frac{1}{x-y}$= $\frac{2}{3}$

$\Rightarrow$ x - y = $\frac{3}{2}$ ... (6)

Adding equations (5) and (6)

$\Rightarrow$ x+y+x-y=$\frac{15}{4}$+$\frac{3}{2}$

$\Rightarrow$ 2x = $\frac{15+6}{4}$

$\Rightarrow$ 2x = $\frac{21}{4}$

$\Rightarrow$ x =$\frac{21}{8}$

Substitute x =$\frac{21}{8}$ in equation (5)

$\Rightarrow$ $\frac{21}{8}$+y=$\frac{15}{4}$

$\Rightarrow$ y = $\frac{15}{4}$-$\frac{21}{8}$

$\Rightarrow$ y =$\frac{30-21}{8}$

$\Rightarrow$ y = $\frac{9}{8}$

Hence $x=\frac{21}{8},y=\frac{9}{8}$