Solve: 3/(x+y)+2/(x-y)=3 and 2/(x+y)+3/(x-y)=11/3
$\frac{3}{{\mathrm{x}} + {\mathrm{y}}} + \frac{2}{{\mathrm{x}} – {\mathrm{y}}} = 3$ —(i)
$\frac{2}{{\mathrm{x}} + {\mathrm{y}}} + \frac{3}{{\mathrm{x}} – {\mathrm{y}}} = \frac{11}{3}$ —(ii)
Let,x+y=a and x-y=b then the given equations become
$\frac{3}{{\mathrm{a}}} + \frac{2}{{\mathrm{b}}} = 3$ —-(iii)
and $\frac{2}{{\mathrm{a}}} + \frac{3}{{\mathrm{b}}} = \frac{11}{3}$ —-(iv)
Now Multiplying Equation (iii) by 3 and Equation (iv) by 2 we get,
$\frac{9}{{\mathrm{a}}} + \frac{6}{{\mathrm{b}}} = 9$ —(v)
and $\frac{4}{{\mathrm{a}}} + \frac{6}{{\mathrm{b}}} = \frac{22}{3}$ —(vi)
Now Subtracting Equations (v) and (vi) we get,
$\frac{9}{{\mathrm{a}}} + \frac{6}{{\mathrm{b}}} – \frac{4}{{\mathrm{a}}} – \frac{6}{{\mathrm{b}}}$ = 9-$\frac{22}{3}$
⟹ $\frac{9}{{\mathrm{a}}} – \frac{4}{{\mathrm{a}}}$ = $\frac{27 – 22}{3}$
⟹ $\frac{5}{{\mathrm{a}}}$ =$\frac{5}{3}$
⟹ a = 3
⟹ x+y = 3 —(vii)
Now Substituting thevalue of a in equation (iii) we get,
$\frac{3}{{\mathrm{a}}} + \frac{2}{{\mathrm{b}}} = 3$
$\Rightarrow \frac{3}{3} + \frac{2}{{\mathrm{b}}} = 3$
$\Rightarrow 1 + \frac{2}{{\mathrm{b}}} = 3$
$\Rightarrow \frac{2}{{\mathrm{b}}} = 3 – 1$
$\Rightarrow \frac{2}{{\mathrm{b}}} = 2$
$\Rightarrow b = \frac{2}{2}$
$\Rightarrow b = 1$
⟹ x-y = 1 —-(viii)
adding equations (vii) and (viii) we get,
x+y+x-y=3+1
⟹ 2x=4
⟹ x =$\frac{4}{2}$
⟹ x = 2
From Equation (viii) we get,
2-y = 1
⟹ 2-1=y
⟹ y =1
Therefore, Required Solutions are x =2 and y= 1 [Ans]