Question
Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
✓
Answer
$\frac{10}{x+y}+\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $\frac{1}{x+y}=m$ and $\frac{1}{x-y}=n$
$10 m+2 n=4$
$15 m-5 n=-2 \ldots II )$
Multiply Eq. $I$ by $5$ and Eq.$II$ by $2$
$50 m+10 n=20$
$\frac{30 m-10 n=-4}{80 m=16}$
$m=\frac{16}{80}$
$m =\frac{1}{5}$
Substituting $m =\frac{1}{5}$ in Eq. I
$10 \times \frac{1}{5}+2 n =4$
$2+2 n =4$
$2 n =4-2$
$2 n =2$
$n =\frac{2}{2}$
$n =1$
$\therefore m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5$
$\therefore n =\frac{1}{ x - y } \Rightarrow \frac{1}{ x - y }=1 \Rightarrow x - y =1$
Now, equating Eq. $III$ and $IV$
$x+y=5$
$x-y=1$
$2 x=6$
$x=\frac{6}{2}$
$x=3$
Subsituting value of $x=3$ in Eq. III
$3+y=5$
$y=5-3$
$y=2$
Hence $(x, y)=(3,2)$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $\frac{1}{x+y}=m$ and $\frac{1}{x-y}=n$
$10 m+2 n=4$
$15 m-5 n=-2 \ldots II )$
Multiply Eq. $I$ by $5$ and Eq.$II$ by $2$
$50 m+10 n=20$
$\frac{30 m-10 n=-4}{80 m=16}$
$m=\frac{16}{80}$
$m =\frac{1}{5}$
Substituting $m =\frac{1}{5}$ in Eq. I
$10 \times \frac{1}{5}+2 n =4$
$2+2 n =4$
$2 n =4-2$
$2 n =2$
$n =\frac{2}{2}$
$n =1$
$\therefore m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5$
$\therefore n =\frac{1}{ x - y } \Rightarrow \frac{1}{ x - y }=1 \Rightarrow x - y =1$
Now, equating Eq. $III$ and $IV$
$x+y=5$
$x-y=1$
$2 x=6$
$x=\frac{6}{2}$
$x=3$
Subsituting value of $x=3$ in Eq. III
$3+y=5$
$y=5-3$
$y=2$
Hence $(x, y)=(3,2)$
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