Question

Find value of $y$,$\frac{57}{x+y}+\frac{6}{x-y}=5$, $\frac{38}{x+y}+\frac{21}{x-y}=9$.

Answer 1

$\frac{57}{x+y}+\frac{6}{x-y}=5$

and ,

$\frac{38}{x+y}+\frac{21}{x-y}=9$

Let$,x+y=\frac{1}{p}$

and$x-y=\frac{1}{q}\therefore 57p+6q=5⟶\left(i\right)38p+21q=9⟶\left(ii\right)$

Multiplying $\left(i\right)$ by $7$ and $\left(ii\right)$ by $2$

$399p+42q=3576p+42q=18$

On substraction : $323p=17p=\frac{17}{323}=\frac{1}{19}$

Putting value of $p$ equation$⟶\left(i\right)$

$3+6q=56q=2q=\frac{1}{3}\therefore x+y=19⟶\left(iii\right)\{\because x+y=\frac{1}{p}\}$

and$,x-y=3⟶\left(iv\right)\{\because x-y=\frac{1}{q}\}$

On adding : $2x=22x=11$

Putting value of $x$ in equation $⟶\left(iii\right)$

$y+11=19$

$y=8$