Question
Solve :$\frac{20}{x+y}+\frac{3}{x-y}=7 ,\frac{8}{x-y}-\frac{15}{x+y}=5$
✓
Answer
$\frac{20}{x+y}+\frac{3}{x-y}=7\ldots . .(1)$
$\frac{8}{x-y}-\frac{15}{x+y}=5\ldots . . .(2)$
Multiplying equation no $(1)$ by $8$ and $(2)$ by $3 .$
$\frac{160}{x+y}+\frac{24}{x-y}=56\ldots . . .(3)$
$-\frac{45}{x+y}+\frac{24}{x-y}=15\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$\frac{160}{x+y}+\frac{24}{x-y}=56$
$-\frac{45}{x+y}+\frac{24}{x-y}=15$
$\frac{205}{x+y}= 41$
$\frac{205}{x+y}=41$
$x+y=5\ldots . . . .(5)$
From $(1)$
$ \frac{20}{5}+\frac{3}{x-y}=7$
$ \frac{3}{x-y}=3$
$x-y=1\ldots . . .(6)$
Adding equation $(5)$ and $(6)$
$x+y=5$
$+ x-y=1$
$2 x=6$
$x=3$
From $(5)$
$3+y=5$
$ y=5-3$
$ y=2$
$\frac{8}{x-y}-\frac{15}{x+y}=5\ldots . . .(2)$
Multiplying equation no $(1)$ by $8$ and $(2)$ by $3 .$
$\frac{160}{x+y}+\frac{24}{x-y}=56\ldots . . .(3)$
$-\frac{45}{x+y}+\frac{24}{x-y}=15\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$\frac{160}{x+y}+\frac{24}{x-y}=56$
$-\frac{45}{x+y}+\frac{24}{x-y}=15$
$\frac{205}{x+y}= 41$
$\frac{205}{x+y}=41$
$x+y=5\ldots . . . .(5)$
From $(1)$
$ \frac{20}{5}+\frac{3}{x-y}=7$
$ \frac{3}{x-y}=3$
$x-y=1\ldots . . .(6)$
Adding equation $(5)$ and $(6)$
$x+y=5$
$+ x-y=1$
$2 x=6$
$x=3$
From $(5)$
$3+y=5$
$ y=5-3$
$ y=2$
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