System of non-linear equations

**Phire** · May 4th 2009, 03:30 PM

I read this problem on a site:

$x^2-xy+y^2=21$
$x^2+xy-8y^2=0$

I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

(x+4y)(x–2y)

**Prove It** · May 4th 2009, 03:47 PM

Originally Posted by Phire

I read this problem on a site:

$x^2-xy+y^2=21$
$x^2+2xy-8y^2=0$

I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

(x+4y)(x–2y)

The Quadratic Formula looks to be easiest, though Completing the Square would be OK too.

Do it on the 2nd equation.

If $x^2 + 2xy - 8y^2 = 0$ then

$x = \frac{-2y \pm \sqrt{(2y)^2 - 4(1)(-8y^2)}}{2(1)}$

$= \frac{-2y \pm \sqrt{36y^2}}{2}$

$= \frac{-2y \pm 6y}{2}$

$= \frac{4y}{2}\textrm{ or }\frac{-8y}{2}$

$= 2y\textrm{ or }-4y$ .

Substitute these values into the first equation and you should be able to solve for y, and then back-substitute to find x.

**skeeter** · May 4th 2009, 03:48 PM

Originally Posted by Phire

I read this problem on a site:

$x^2-xy+y^2=21$
$x^+2xy-8y^2=0$

I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

(x+4y)(x–2y)

$x^2 + 2xy - 8y^2 = 0<br />$

$(x+4y)(x-2y) = 0$

$x = -4y$ or $x = 2y$

go back to the first equation ...

$x^2 - xy + y^2 = 21$

sub in $-4y$ for $x$ ...

$16y^2 + 4y^2 + y^2 = 21$

$21y^2 = 21$

$y = \pm 1$ ... $x = \mp 4$

sub in $2y$ for $x$ ...

$4y^2 - 2y^2 + y^2 = 21$

$3y^2 = 21$

$y = \pm \sqrt{7}$ ... $x = \pm 2\sqrt{7}$

**Phire** · May 4th 2009, 04:48 PM

Thanks, but can you guys tell me how you were able to factor the 2nd equation? I can't seem to do that. I tried:

$x^2 + 2xy-8y^2$

$x(x+2y)...$ Already don't know what to do

So then I try:

$x^2 +xy+xy-8y^2$

$x(x+y)y(x+8y)$ No good...

**mr fantastic** · May 4th 2009, 08:18 PM

Originally Posted by Phire

Thanks, but can you guys tell me how you were able to factor the 2nd equation? I can't seem to do that. I tried:

$x^2 + 2xy-8y^2$

$x(x+2y)...$ Already don't know what to do

So then I try:

$x^2 +xy+xy-8y^2$

$x(x+y)y(x+8y)$ No good...

Post #2 tells you how to find the roots. And if you know the roots then you know the factors. But getting the roots is all you need.

Alternatively:

$x^2 + 2xy - 8y^2 = (x^2 - 2xy) + (4xy - 8y^2) = x(x - 2y) + 4y(x - 2y) = (x + 4y)(x - 2y)$ .

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