Question

Solve the following pair of equation by reducing them to a pair of linear equation:-
i) $\frac{1}{2x}+\frac{1}{3y}=2$
ii) $\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

Answer 1

Part 1^st

Given that

$\frac{1}{2x}+\frac{1}{3y}=2$ ……(1)

$\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$ …..(2)

Put $u=\frac{1}{x},v=\frac{1}{y}$

We get

$\frac{u}{2}+\frac{v}{3}=2$ …. (3)

$\frac{u}{3}+\frac{v}{2}=\frac{13}{6}$ …..(4)

Multiply by $\frac{1}{3}$ in equation (3)and $\frac{1}{2}$ in equation (4)

then subtract from equation (3) to (4) we get

$v(\frac{1}{9}-\frac{1}{4})=\frac{2}{3}-\frac{13}{12}$

$v(-\frac{5}{36})=\frac{-11}{36}$

$v=\frac{11}{5}$

The value of v put in equation (3)

$\frac{u}{2}+\frac{1}{3}\times \frac{11}{5}=2$

$u=\frac{38}{15}$

$u=\frac{1}{x}=\frac{38}{15}$

$x=\frac{15}{38}$

$v=\frac{1}{y}=\frac{11}{5}$

$y=\frac{5}{11}$