Solve for x and y : $\frac{44}{x+y}+\frac{30}{x-y}=10$, $\frac{55}{x+y}-\frac{40}{x-y}=13$

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Solve for x and y :
$\frac{44}{x+y}+\frac{30}{x-y}=10$ ,
$\frac{55}{x+y}-\frac{40}{x-y}=13$

Solve for x and y :
$\frac{44}{x+y}+\frac{30}{x-y}=10$ ,
$\frac{55}{x+y}-\frac{40}{x-y}=13$

Solution:

The given eq. are
$\frac{44}{x+y}+\frac{30}{x-y}=10 \dots \dots(i)$
$\frac{55}{x+y}-\frac{40}{x-y}=13\dots \dots (ii)$
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , we get:

$44 u+30 v=10\dots \dots(iii)$
$55 u+40 v=13 \dots \dots(iv)$
On multiplying equation(iii) by 4 and equation(iv) by 3 , we obtain:
$\begin{array}{ll} 176 \mathrm{u}+120 \mathrm{v}=40 & \ldots \ldots(\mathrm{v}) \\ 165 \mathrm{u}+120 \mathrm{v}=39 & \ldots \ldots(\mathrm{vi}) \end{array}$
On subtracting equation(vi) and equation(v), we obtain:
$\begin{array}{l} 11 \mathrm{u}=1 \\ \Rightarrow \mathrm{u}=\frac{1}{11} \\ \Rightarrow \frac{1}{x+y}=\frac{1}{11} \Rightarrow \mathrm{x}+\mathrm{y}=11\dots \dots(vii) \end{array}$
On substituting $\mathrm{u}=\frac{1}{11}$ in equation(iii), we obtain:
$\begin{array}{l} 4+30 \mathrm{v}=10 \\ \Rightarrow 30 \mathrm{v}=6 \\ \Rightarrow \mathrm{v}=\frac{6}{30}=\frac{1}{5} \\ \Rightarrow \frac{1}{x-y}=\frac{1}{5} \Rightarrow \mathrm{x}-\mathrm{y}=5\dots \dots(viii) \end{array}$
On adding equation(vii) and equation(viii), we obtain
$\begin{array}{l} 2 \mathrm{x}=16 \\ \Rightarrow \mathrm{x}=8 \end{array}$
On substituting $x=8$ in equation(vii), we obtain:
$\begin{array}{l} 8+y=11 \\ \Rightarrow y=11-8=3 \end{array}$
As a result, the required solution is $\mathrm{x}=8$ and $\mathrm{y}=3$ .