Question

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

Answer 1

$\frac{1}{2x}+\frac{1}{3y}=2$
$\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$
Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$, then the equations changes as below:
$\frac{p}{2}+\frac{q}{3}=2$
$\Rightarrow 3p+2q-12=0$ ... (i)
$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$
$\Rightarrow 2p+3q-13=0$ ... (ii)

Compare the given equations with

$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$.

By cross-multiplication method, we get

$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$
$\frac{p}{-26-(-36)}=\frac{q}{-24-(-39)}=\frac{1}{9-4}$
$\frac{p}{10}=\frac{q}{15}=\frac{1}{5}$
$\frac{p}{10}=\frac{1}{5}$ and $\frac{q}{15}=\frac{1}{5}$
p=2 and q=3
$\frac{1}{x}=2$ and $\frac{1}{y}=3$
Hence, $x=\frac{1}{2}$ and $y=\frac{1}{3}$