3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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:

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

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

$\endgroup$
  • $\begingroup$ of course I can , thank you a lot $\endgroup$ – user348056 Jun 15 '16 at 21:29
  • $\begingroup$ You're very welcome. $\endgroup$ – Ahmed S. Attaalla Jun 15 '16 at 21:30
  • $\begingroup$ Shouldn't it be 32 instead of 8 on the RHS, line 2? $\endgroup$ – Bayesian Jun 15 '16 at 21:33
  • $\begingroup$ actually , after trying . also , I can't $\endgroup$ – user348056 Jun 15 '16 at 21:47
  • $\begingroup$ can you put your full sol please $\endgroup$ – user348056 Jun 15 '16 at 21:48
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ How did you factor it? Is there some method to factoring something of this sort. $\endgroup$ – Ahmed S. Attaalla Jun 15 '16 at 21:44
  • 1
    $\begingroup$ If it factors into two polynomials with integer coefficients, then obviously the product of the two constant terms has to equal the constant term, and the product of the two leading terms has to equal the leading term. So you work through the possibilities. You start with a skeleton $(3s^2+hx\pm7)(7s^2+ks\pm3)$ and see whether it works. Then move to another one if it fails. With practice the process is fairly quick. Of course it may split as a linear factor. But for that it is often quicker to try the factors of the constant term directly eg is +1 a root etc. $\endgroup$ – almagest Jun 15 '16 at 21:49
  • $\begingroup$ I there is something wrong $\endgroup$ – user348056 Jun 15 '16 at 22:19
  • $\begingroup$ It may originally look different from my previous answer, but through conjugates I can confirm its the same.@user348056 $\endgroup$ – Ahmed S. Attaalla Jun 16 '16 at 0:50
  • $\begingroup$ I should've used the conjugate when I got $x+y$ would have simplified calculations very much @user348056. $\endgroup$ – Ahmed S. Attaalla Jun 16 '16 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.