Question

Solve the following pair of equations for $x$ and $y\frac{15}{x-y}+\frac{22}{x+y}=5$ & $\frac{40}{x-y}+\frac{55}{x+y}=13x\ne y$ and $x\ne -y$.

Answer 1

$\frac{15}{x-y}+\frac{22}{x+y}=5$ ......$\left(1\right)$

$\frac{40}{x-y}+\frac{55}{x+y}=13$ ......$\left(2\right)$

Let $u=\frac{1}{x-y}$ and $v=\frac{1}{x+y}$ then equations $\left(1\right)$ and $\left(2\right)$ becomes,

$15u+22v=5$ .....$\left(3\right)$

$40u+55v=13$ .....$\left(4\right)$

$40\times$Eqn$\left(3\right)-15\times$Eqn$\left(4\right)$

$600u+880v-600u-825v=200-195$

$\Rightarrow 55v=5$

$\therefore v=\frac{5}{55}=\frac{1}{11}$

$\Rightarrow v=\frac{1}{x+y}=\frac{1}{11}$

$\therefore x+y=11$ ......$\left(5\right)$

Substitute $v=\frac{1}{11}$ in eqn$\left(3\right)$ we get

$15u+22\times \frac{1}{11}=5$

$\Rightarrow 15u=5-2=3$

$\Rightarrow u=\frac{3}{15}=\frac{1}{5}$

$\Rightarrow u=\frac{1}{x-y}=\frac{1}{5}$

$\therefore x-y=5$ ........$\left(6\right)$

Adding eqns$\left(5\right)$ and $\left(6\right)$ we get

$x+y+x-y=11+5$

$\Rightarrow 2x=16$ or $x=\frac{16}{2}=8$

$\therefore x=8$

Substitute $x=8$ in $\left(5\right)$ we get

$x+y=11\Rightarrow y=11-x=11-8=3$

$\therefore y=11$

Hence $x=8,y=11$