How can I solve this hard system of equations?

Question

Solve the system below

\begin{align} &\sqrt {3x} \left( 1+\frac {1}{x+y} \right) =2\\ &\sqrt {7y} \left( 1-\frac{1}{x+y} \right) =4\sqrt{2} \end{align}

Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I hope you can help me in approaching this problem.

Answer 1

Square both equations. Then let $u=\frac{1}{x+y}$. Note then $x+y=\frac{1}{u}$

$$3x(1+\frac{1}{x+y})^2=4$$

$$7y(1-\frac{1}{x+y})^2=32$$

So.

$$x(1+u)^2=\frac{4}{3}$$

$$y(1-u)^2=\frac{32}{7}$$

And for $u \neq 1$

$$x=\frac{4}{3}(1+u)^{-2}$$

$$y=\frac{32}{7}(1-u)^{-2}$$

Adding both equations we have:

$$\frac{1}{u}=\frac{4}{3(1+u)^2}+\frac{32}{7(1-u)^2}$$

$$\frac{1}{u}=\frac{28(1-u)^2+96(1+u)^2}{21(1-u)^2(1+u)^2}$$

Cross multiplication, and moving terms to one side yields:

$$28u(1-u)^2+96u(1+u)^2-21(1-u)^2(1+u)^2=0$$

$$-21u^4+124u^3+178u^2+124u-21=0$$

This factors:

$$-(3-22u+3u^2)(7+10u+7u^2)=0$$

So:

$$u=\frac{11 \pm 4\sqrt{7}}{3}$$

$$x+y=\frac{3}{11 \pm 4\sqrt{7}}$$

Back substitution yields:

$$\sqrt{3x}(\frac{14 \pm 4\sqrt{7}}{3})=2$$

$$x=\frac{12}{(14 \pm 4\sqrt{7})^2}$$

$$y=\frac{3}{11 \pm 4\sqrt{7}}-\frac{12}{(14 \pm 4\sqrt{7})^2}$$

We introduced an extraneous solution when squaring so checking the possible solutions, the final solution is:

$$x=\frac{12}{(14-4\sqrt{7})^2}$$

$$y=\frac{3}{11-4\sqrt{7}}-\frac{12}{(14- 4\sqrt{7})^2}$$

If you wish, from the second equation, you can express $y$ as:

$$y=\frac{288}{7(4\sqrt{7}-8)^2}$$

In fact through the method of multiplying by conjugates we may get:

$$x=\frac{1}{21}(11+4\sqrt{7})$$

$$y=\frac{2}{7}(11+4\sqrt{7})$$

Answer 2

Squaring 1st equation you have $3x(x+y+1)^2=4(x+y)^2$, squaring second you have $7y(x+y-1)^2=32(x+y)^2$. Put $s=x+y,y=s-x$ and we get $3s^2x+6sx+3x-4s^2=0,7s^3-7s^2x+14sx-46s^2+7s=0$. The first of these gives $x=\frac{4s^2}{3s^2+6s+3}$.

Substituting in the second we get $$s(21s^4-124s^3-178s^2-124s+21)=0$$ or $$s(3s^2-22s+3)(7s^2+10s+7)=0$$ It is clear from the original equations that $s\ne0$ and $7s^2+10s+7=0$ has no real solutions, so we have $s=\frac{1}{3}(11\pm4\sqrt7)$.

Using the equation for $x$ we get $x=\frac{1}{21}(11+4\sqrt7),y=\frac{2}{7}(11+4\sqrt7)$ or $x=\frac{1}{21}(11-4\sqrt7),y=\frac{22}{7}(11-\frac{8}{\sqrt7})$. But checking with the original equations the second solution fails (because $s-1$ is negative).