Question

Solve for x and y:
$$\begin{gathered} \frac{1}{\left(3x+y\right)}+\frac{1}{\left(3x-y\right)}=\frac{3}{4} \\ \frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=\frac{-1}{8} \end{gathered}$$

Answer 1

The given equations are
$$\begin{gathered} \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}.....\left(i\right) \\ \frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=-\frac{1}{8} \\ \frac{1}{3x+y}-\frac{1}{3x-y}=-\frac{1}{4}\left(\mathrm{Multiplying}\mathrm{by}2\right).....\left(ii\right) \end{gathered}$$
Substituting $\frac{1}{3x+y}=u\mathrm{and}\frac{1}{3x-y}=v$ in (i) and (ii), we get
$$\begin{gathered} u+v=\frac{3}{4}.....\left(iii\right) \\ u-v=-\frac{1}{4}.....\left(iv\right) \end{gathered}$$
Adding (iii) and (iv), we get
$$\begin{gathered} 2u=\frac{1}{2} \\ \Rightarrow u=\frac{1}{4} \\ \Rightarrow 3x+y=4\left(\because \frac{1}{3x+y}=u\right).....\left(v\right) \end{gathered}$$
Now, substituting $u=\frac{1}{4}$ in (iii), we get
$$\begin{gathered} \frac{1}{4}+v=\frac{3}{4} \\ v=\frac{3}{4}-\frac{1}{4}\Rightarrow v=\frac{1}{2} \\ \Rightarrow 3x-y=2\left(\because \frac{1}{3x-y}=v\right).....\left(vi\right) \end{gathered}$$
Adding (v) and (vi), we get
$6x=6\Rightarrow x=1$
Substituting x = 1 in (v), we have
$3+y=4\Rightarrow y=1$
Hence, x = 1 and y = 1.