Question

$\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=-\frac{3}{2},\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60,}\mathrm{where}x+2y\ne 0\mathrm{and}3x-2y\ne 0.$

Answer 1

The given equations are:
$\frac{1}{2\left(x+2y\right)}+\frac{5}{3\left(3x-2y\right)}=-\frac{3}{2}$ ...(i)
$\frac{5}{4\left(x+2y\right)}-\frac{3}{5\left(3x-2y\right)}=\frac{61}{60}$ ...(ii)
Putting $\frac{1}{x+2y}=u$ and $\frac{1}{3x-2y}=v$ , we get:
$\frac{1}{2}u+\frac{5}{3}v=-\frac{3}{2}$ ...(iii)
$\frac{5}{4}u-\frac{3}{5}v=\frac{61}{60}$ ...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9 ...(v)
$25u-12v=\frac{61}{3}$ ...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54 ...(vii)
$125u-60v=\frac{305}{3}$ ...(viii)
On adding (vii) and (viii), we get:
$143u=\frac{305}{3}-54=\frac{305-162}{3}=\frac{143}{3}$
$\Rightarrow u=\frac{1}{3}=\frac{1}{x+2y}$
⇒ x + 2y = 3 ...(ix)
On substituting $u=\frac{1}{3}$ in (v), we get:
1 + 10v = −9
⇒ 10v = −10
⇒ v = −1
$\Rightarrow \frac{1}{3x-2y}=-1\Rightarrow 3x-2y=-1$ ...(x)
On adding (ix) and (x), we get:
4x = 2
⇒ $x=\frac{1}{2}$
On substituting $x=\frac{1}{2}$ in (x), we get:
$$\begin{gathered} \frac{3}{2}-2y=-1 \\ \Rightarrow 2y=\left(\frac{3}{2}+1\right)=\frac{5}{2} \\ \Rightarrow y=\frac{5}{4} \end{gathered}$$
Hence, the required solution is $x=\frac{1}{2}\mathrm{and}\mathrm{y}=\frac{5}{4}$.