Solve the following systems of equations:
Question:

Solve the following systems of equations:

$\frac{44}{x+y}+\frac{30}{x-y}=10$

$\frac{55}{x+y}+\frac{40}{x-y}=13$

Solution:

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 $y=3$