Question

Solve for x and y:
$$\begin{gathered} \frac{2}{\left(3x+2y\right)}+\frac{3}{\left(3x-2y\right)}=\frac{17}{5} \\ \frac{5}{\left(3x+2y\right)}+\frac{1}{\left(3x-2y\right)}=2 \end{gathered}$$

Answer 1

The given equations are
$$\begin{gathered} \frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}.....\left(i\right) \\ \frac{5}{3x+2y}+\frac{1}{3x-2y}=2.....\left(ii\right) \end{gathered}$$
Substituting $\frac{1}{3x+2y}=u\mathrm{and}\frac{1}{3x-2y}=v$ in (i) and (ii), we get
$$\begin{gathered} 2u+3v=\frac{17}{5}.....\left(iii\right) \\ 5u+v=2.....\left(iv\right) \end{gathered}$$
Multiplying (iv) by 3 and subtracting from(iii), we get
$$\begin{gathered} 2u-15u=\frac{17}{5}-6 \\ \Rightarrow -13u=\frac{-13}{5}\Rightarrow u=\frac{1}{5} \\ \Rightarrow 3x+2y=5\left(\because \frac{1}{3x+2y}=u\right).....\left(v\right) \end{gathered}$$
Now, substituting $u=\frac{1}{5}$ in (iv), we get
$$\begin{gathered} 1+v=2\Rightarrow v=1 \\ \Rightarrow 3x-2y=1\left(\because \frac{1}{3x-2y}=v\right).....\left(vi\right) \end{gathered}$$
Adding (v) and (vi), we get
$6x=6\Rightarrow x=1$
Substituting x = 1 in (v), we have
$3+2y=5\Rightarrow y=1$
Hence, x = 1 and y = 1.