Question

Solve the following system of equations
$\frac{6}{x+y}=\frac{7}{x-y}+3,\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$ where $x+y\ne 0andx-y\ne 0$

Answer 1

$\frac{6}{x+y}=\frac{7}{x-y}+3$ ......(i)

Put x + y = a, x - y = b

Now the given equation reduces to

$\frac{6}{a}=\frac{7}{b}+3$ ........(ii)

Another equation is $\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$ .......(iii)

$\frac{1}{2a}=\frac{1}{3b}$

or a = $\frac{3b}{2}$ ........(iv)

Put (iv) in (ii)

$\frac{6}{3b}\times 2=\frac{7+3b}{b}$

$4=7+3b$

$b=-1$ ......(v)

Put (v) in (iv)

$a=-\frac{3}{2}$

From our four assumptions

a = x + y = $-\frac{3}{2}$

$b=x-y=-1$

Adding the two equations given above

$2x=-\frac{5}{2}\Rightarrow x=-\frac{5}{4}$

$\therefore y=-\frac{5}{4}+1\Rightarrow y=-\frac{1}{4}$