Solve for $x$ and $y: \frac{3}{x+y}+\frac{2}{x-y}=2, \frac{9}{x+y}-\frac{4}{x-y}=1$

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Solve for $x$ and $y: \frac{3}{x+y}+\frac{2}{x-y}=2, \frac{9}{x+y}-\frac{4}{x-y}=1$

Solution:

The given system is
$\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2\dots\dots(i) \\ \frac{9}{x+y}-\frac{4}{x-y}=1\dots \dots(ii) \end{array}$
Substituting $\frac{1}{x+y}=\mathrm{u}$ and $\frac{1}{x-y}=\mathrm{v}$ in equation(i) and equation(ii), the given equations are changed to
$\begin{array}{l} 3 u+2 v=2\dots \dots(iii) \\ 9 u-4 v=1\dots \dots(iv) \end{array}$
On multiplying equation(i) by 2 and adding it with equation(ii), we obtain
$15 \mathrm{u}=4+1 \Rightarrow \mathrm{u}=\frac{1}{3}$
Multiplying equation(i) by 3 and subtracting equation(ii) from it, we obtain
$6 \mathrm{u}+4 \mathrm{v}=6-1 \Rightarrow \mathrm{u}=\frac{5}{10}=\frac{1}{2}$
So
$\begin{array}{l} x+y=3\dots \dots(v) \\ x-y=2\dots \dots(vi) \end{array}$
Now, adding equation(v) and equation(vi) we get
$2 x=5 \Rightarrow x=\frac{5}{2}$
On substituting $\mathrm{x}=\frac{5}{2}$ in equation(v), we have
$\frac{5}{2}+y=3 \Rightarrow y=3-\frac{5}{2}=\frac{1}{2}$
As a result, $x=\frac{5}{2}$ and $\mathrm{y}=\frac{1}{2}$ .