Question

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

Given:

The pair of linear equations are,

$\frac{1}{2x}+\frac{1}{3y}=2$
$\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$
Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$, then the equations changes as below:
$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow \frac{3p+2q}{6}=2$
$\Rightarrow 3p+2q-12=0$ ... (i)
$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow \frac{2p+3q}{6}=\frac{13}{6}$
$\Rightarrow 2p+3q-13=0$ ... (ii)
Multiply eq (i) by $2$

$\Rightarrow 6p+4q-24=0$ ----(iii)

Multiply eq (ii) by $3$

$\Rightarrow 6p+9q-39=0$ ---(iv)

Lets do (iii)-(iv)

$\Rightarrow 6p+4q-24-(6p+9q-39)=0$

$\Rightarrow 6p+4q-24-6p-9q+39=0$

$\Rightarrow -5q+15=0$

$\Rightarrow -5q=-15$

$\Rightarrow 5q=15$

$\Rightarrow q=\frac{15}{5}$

$\Rightarrow q=3$

Now, $\frac{1}{y}=q$

$\Rightarrow y=\frac{1}{q}=\frac{1}{3}$---(v)

Lets substitute $q=3$ in $6p+9q-39=0$

$\Rightarrow 6p+9\left(3\right)-39=0$

$\Rightarrow 6p+27-39=0$

$\Rightarrow 6p-12=0$

$\Rightarrow 6p=12$

$\Rightarrow p=\frac{12}{6}$

$\Rightarrow p=2$

Now, $\frac{1}{x}=p$

$\Rightarrow x=\frac{1}{p}=\frac{1}{2}$ ---(vi)

From (v) and (vi), $x=\frac{1}{2}$ and $y=\frac{1}{3}$