Solve the given simultaneous equations

[chwala]

Homework Statement

$\dfrac{x+2}{y-4} +\dfrac{2(y-4)}{x+2} + 3=0$

$x-y=3$

Relevant Equations

knowledge of simultaneous equations

$\dfrac{x+2}{y-4} +\dfrac{2(y-4)}{x+2} + 3=0$
$x-y=3$

My approach, will call it brute force, i have

$y=x-3$
then,

$\dfrac{(x+2)^2+2(y-4)^2}{(x+2)(y-4)}=-3$
$(x+2)^2+2(y-4)^2=-3(x+2)(y-4)$
$(x+2)^2+2(x-7)^2=-3(x+2)(x-7)$
$x^2+4x+4+2(x^2-14x+49)+3(x^2-5x-14)=0$
$6x^2-39x+60=0$
$x_1=4, y_1=1$ and $x_2=2.5, y_2=-0.5$

there could be a much better and simpler approach that prompted my post.

cheers man!

 

[jedishrfu]

That's how most people would solve it.

 

[DaveE]

I don't know if it's really easier or much different, but you can use the symmetry and a substitution
$u = \frac{(x+2)}{(y-4)}$ and solve $u^2 + 3u +2 =0$ with $u= -1, -2$
switching back to x, y gives $x = 6 - 2y$ or $x = 2-y$
This can then be combined with the given constraint $x-y=3$ to get the two solutions.

PS: Yes, I left out some intermediate steps.

 

[Gavran]

$\frac{x+2}{y-4}+\frac{2(y-4)}{x+2}+3=0$
$\frac{x+2+2(y-4)}{y-4}+\frac{2(y-4)+x+2}{x+2}=0 \Rightarrow (x+2y-6)(x+y-2)=0$

and there are two systems of equations with two different solutions.
The first one is
$x+2y=6$
$x-y=3$
and the second one is
$x+y=2$
$x-y=3$.

 

[Mark44]

It should be stated implicitly, because of the first equation, that $y \ne 4$ and $x \ne -2$. Then, due to the second equation, which is equivalent to y = x -3, we have to disallow x = 7 and y = -5.

 

[pasmith]

You cna simplify the algebra by setting $a = x + 2$, $b = y - 4$ so that $$\begin{split} \frac{a}{b} + 2 \frac{b}{a} + 3 &= 0 \\ a - b &= 9 \end{split}$$ Then multiplying the first by $ab$ gives $$a^2 + 3ab + 2b^2 = (a + b)(a + 2b) = 0$$ and thus we have two solutions, $$\left.\begin{array}{rcl} a + b &=& 0 \\ a - b &=& 9 \end{array} \right\}\quad \mbox{and}\quad \left\{\begin{array}{rcl} a + 2b &=& 0 \\ a - b &=& 9 \end{array}\right..$$