Question

If $\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}{2},\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$ where $x+2y\ne 0and3x-2y\ne 0$ Then what will be the values of x and y?

Answer 1

Let $\frac{1}{x+2y}=U,\frac{1}{3x-2y}=V\Rightarrow \frac{U}{2}+\frac{5V}{3}=-\frac{3}{2}$

$\Rightarrow 3U+10V=-\mathrm{9.........}\left(i\right)$

$\Rightarrow \frac{5U}{4}-\frac{3V}{5}=\frac{61}{60}\Rightarrow 75U-36V=\mathrm{61.........}\left(ii\right)$

Equation (I) is multiplied by 25

75U + 250V = -225

75U - 36V = 61

- + -

Subtracting 286V = - 286 $V=-1,U=\frac{1}{3}$

$\therefore \frac{1}{x+2y}=\frac{1}{3},\frac{1}{3x-2y}=-1$ x + 2y = 3, -3x + 2y = 1

x + 2y = 3

-3x + 2y = 1

+ - - Subtracting 4x = 2 So $x=\frac{1}{2},y=\frac{5}{4}$