Question

Solve: $\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2}$ and

$\frac{7}{2x+3y}+\frac{4}{3x-2y}=2$, where $2x+3y\ne 0$ and $3x-2y\ne 0$.

Answer 1

Let $\frac{1}{2x+3y}=u$ and $\frac{1}{3x-2y}=v$. Then, the given system of equations becomes

$\frac{1}{2}u+\frac{12}{7}v=\frac{1}{2}\Rightarrow 7u+24v=7$ ..(i)

and, $7u+4v=2$ .(ii)

Substracting equation (ii) from equation (i), we get

$20v=5\Rightarrow v=\frac{1}{4}$

Putting $v=\frac{1}{4}$ in equation (i), we get

$7u+6=7\Rightarrow u=\frac{1}{7}$

Now, $u=\frac{1}{7}\Rightarrow \frac{1}{2x+3y}=\frac{1}{7}\Rightarrow 2x+3y=7$ (iii)

and, $u=\frac{1}{4}\Rightarrow \frac{1}{3x-2y}=\frac{1}{4}\Rightarrow 3x-2y=4$ .(iv)

Multiplying equation (iii) by $2$ and equation (iv) by $3$, we get
$4x+6y=14$ .(v)
$9x-6y=12$ .(vi)

Adding equations (v) and (vi), we get
$13x=26\Rightarrow x=2$

Putting $x=2$ in equation(v), we get
$8+6y=14\Rightarrow y=1$

Hence, $x=2,y=1$ is the solution of the given system of equations.