Question

Solve for x and y:
${\displaystyle \frac{10}{x+y}+\frac{2}{x-y}=4,\frac{15}{x+y}-\frac{9}{x-y}=-2,where x\ne y,x\ne -y.}$

Answer 1

The given equations are
${\displaystyle \frac{10}{x+y}+\frac{2}{x-y}=4}$   ……(i)
${\displaystyle \frac{15}{x+y}-\frac{9}{x-y}=-2}$   ……(ii)
Substituting ${\displaystyle \frac{1}{x+y}=u\text{and}\frac{1}{x-y}}$ = v in (i) and (ii), we get:
10u + 2v = 4 ……..(iii)
15u - 9v = -2 …….(iv)
Multiplying (iii) by 9 and (iv) by 2 and adding, we get:
90u + 30u = 36 –4

⇒120u = 32
${\displaystyle \Rightarrow u=\frac{32}{120}=\frac{4}{15}}$
${\displaystyle \Rightarrow x+y=\frac{15}{4} (\because \frac{1}{x+y}=u)}$  …..(v)
On substituting u = ${\displaystyle \frac{4}{15}}$ in (iii), we get:
${\displaystyle 10\times \frac{4}{15}+2v=4}$
${\displaystyle \frac{8}{3}+2v=4}$
⇒2v = 4 - 83 = 43
${\displaystyle \Rightarrow v=\frac{2}{3}}$
${\displaystyle \Rightarrow x–y=\frac{3}{2} (\because \frac{1}{x-y}=v)}$     …..(vi)
Adding (v) and (vi), we get
${\displaystyle 2x=\frac{15}{4}+\frac{3}{2}\Rightarrow 2x=\frac{21}{4}\Rightarrow x=\frac{21}{8}}$
Substituting x${\displaystyle =\frac{21}{8}}$in (v), we have
${\displaystyle \frac{21}{8}+y=\frac{15}{4}\Rightarrow y=\frac{15}{4}-\frac{21}{8}=\frac{9}{8}}$
Hence, x =${\displaystyle \frac{21}{8}\text{and}y=\frac{9}{8}.}$