Solve: 3/(x+y)+2/(x-y)=3 and 2/(x+y)+3/(x-y)=11/3

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]

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