Question

Solve for x and y:

$\frac{3}{x+y}+\frac{2}{x-y}=2,\frac{9}{x+y}-\frac{4}{x-y}=1.$

Answer 1

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

Let $\frac{1}{x+y}=u$

Let $\frac{1}{x-y}=v$

Soln:

Then the given system of the equation becomes:

$3u+2v=2$…….. (i)

$9u-4v=1$……....(ii)

Multiplying equation (i) by 2 and (ii) by 1

$6u+4v=4$…………… (iii)

$9u-4v=1$…………..... (iv)

Adding equation (iii) and (iv) we get,

$15u=5$

$u=\frac{1}{3}$

multiplying equation (i) by 3

$9u+6v=6$.......(v)

Subtracting equation (iv) from equation (v) , we get

$2v=2-1$

$2v=1$

$v=\frac{1}{2}$

Now,

$\frac{1}{x+y}=u=\frac{1}{3}$

$=x+y=3$……………..(vi )

$\frac{1}{x-y}=v=\frac{1}{2}$

$=x-y=2$…………………..(vii)

Adding equation (vi) and (vii) we get,

$2x=5$

$=x=\frac{5}{2}$

Putting the value of x in equation (vi)

$\frac{5}{2}+y=3$

$y=\frac{1}{2}$

The solutions of the given system of equation are $\frac{5}{2}$ and $\frac{1}{2}$ respectively.