System of nonlinear equations in two variables

Question

Solve: $$\begin{cases}\frac{2}{x+y} - \frac{1}{x-y} = 11\\ \frac{5}{x+y} + \frac{4}{x-y} = 8\end{cases}$$

Please help me out. I solved this sum but my answer is coming as $x=\frac{1}{32}$ and $y=\frac{7}{32}$ but the correct answer should be $x=-\frac{1}{24}$ and $y=\frac{7}{24}$. How is this coming. Please show with steps.

Answer 1

multiplying by $$x+y$$ and $$x-y$$ we get $$2(x-y)-(x+y)=11(x^2-y^2)$$ $$5(x-y)+4(x+y)=8(x^2-y^2)$$ thus we get $$\frac{x-3y}{11}=\frac{9x-y}{8}$$ this can be solved for $x$ or $y$ and you can eliminate one variable

Answer 2

Let $\frac{1}{x+y}=a$ and $\frac{1}{x-y}=b$.

Thus, $2a-b=11$ and $5a+4b=8$, which gives $a=4$ and $b=-3$ or $$x+y=\frac{1}{4}$$ and $$x-y=-\frac{1}{3},$$which after summing and subtraction gives $$(x,y)=\left(\frac{1}{24},\frac{7}{24}\right)$$

Answer 3

Let $a = \dfrac{1}{x+y}$ and $b = \dfrac{1}{x-y}$, note that one should have $x + y \neq 0$ as well as $x - y \neq 0$. Your system can be rewritten as: \begin{cases} 2a - b & = 11 \\ 5a + 4b & = 8 \end{cases} which gives $(a,b) = (4,-3)$. Then solve the system \begin{cases} x + y & = \dfrac{1}{4} \\ x - y & = \dfrac{-1}{3} \end{cases} which gives the desired solution.