Solve the following systems of equations:
$\frac{44}{x+y}+\frac{30}{x-y}=10$
$\frac{55}{x+y}+\frac{40}{x-y}=13$
The given equations are:
$\frac{44}{x+y}+\frac{30}{x-y}=10$
$\frac{55}{x+y}+\frac{40}{x-y}=13$
Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ then equations are
$44 u+30 v=10 \ldots(i)$
$55 u+40 v=13 \ldots($ ii $)$
Multiply equation $(i)$ by 4 and equation $(i i)$ by 3 add both equations, we get
Put the value of $u$ in equation $(i)$, we get
$44 \times \frac{1}{11}+30 v=10$
$\Rightarrow 30 v=6$
$\Rightarrow v=\frac{1}{5}$
Then
$\frac{1}{x+y}=\frac{1}{11}$$\ldots(i i i)$
$\Rightarrow x+y=11$
$\frac{1}{x-y}=\frac{1}{5}$... (iv)
$\Rightarrow x-y=5$
Add both equations, we get
$x+y=11$
Put the value of $x$ in equation (iii) we get
$8 \times 1+y=11$
$\Rightarrow y=3$
Hence the value of $x=8$ and $y=3$
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