Question

If $\frac{3}{x+y}+\frac{2}{x-y}=2\mathrm{and}\frac{9}{x+y}-\frac{4}{x-y}=1$ then
(a) $x=\frac{1}{2},y=\frac{3}{2}$ (b) $x=\frac{5}{2},y=\frac{1}{2}$ (c) $x=\frac{3}{2},y=\frac{1}{2}$ (d) $x=\frac{1}{2},y=\frac{5}{2}$

Answer 1

The given equations are
$$\begin{gathered} \frac{3}{x+y}+\frac{2}{x-y}=2.....\left(i\right) \\ \frac{9}{x+y}-\frac{4}{x-y}=1.....\left(ii\right) \end{gathered}$$
Substituting $\frac{1}{x+y}=u\mathrm{and}\frac{1}{x-y}=v$ in (i) and (ii), the new system becomes
$$\begin{gathered} 3u+2v=2.....\left(iii\right) \\ 9u-4v=1.....\left(iv\right) \end{gathered}$$
Now, multiplying (iii) by 2 and adding it with (iv), we get
$6u+9u=4+1\Rightarrow u=\frac{5}{15}=\frac{1}{3}$
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
$6v+4v=6-1\Rightarrow v=\frac{5}{10}=\frac{1}{2}$
Therefore
$$\begin{gathered} x+y=3.....\left(v\right) \\ x-y=2.....\left(vi\right) \end{gathered}$$
Adding (v) and (vi), we get
$2x=3+2\Rightarrow x=\frac{5}{2}$
Substituting $x=\frac{5}{2}$, in (v), we have
$\frac{5}{2}+y=3\Rightarrow y=3-\frac{5}{2}=\frac{1}{2}$
Thus, $x=\frac{5}{2}\mathrm{and}y=\frac{1}{2}$.
Hence, the correct answer is option (b).