solve the equation

**alternative** · December 14th 2010, 12:14 AM

Hi, can anyone help me solve this

x^2-y^2=4
2x^2+2y^2+3x=4

I multiplyed the first line by -2 and got

-2x^2-2y^2=-8

Then added to the second line where I got

4y^2+3x=-4

So I got the system :

4y^2+3x=-4
x^2-y^2=4

Substitute the second line in the first one then solve the system, but still the result aren't that convinsing that it's solved well.

**Educated** · December 14th 2010, 01:04 AM

I personally like the substitution method. It's easy and always works.

$x^2-y^2=4 \,\,\,\,\, \Longrightarrow \,\,\,\,\, y = \sqrt{x^2 - 4}$

$2x^2+2y^2+3x=4$

Make the substitution of the first equation into the second one.

$2x^2+2(\sqrt{x^2 - 4})^2+3x=4$

$2x^2+2(x^2 - 4)+3x=4$

$4x^2+3x - 12= 0$

Now that it's in the form of a quadratic, it can easily be solved.

Your solution works and gives the correct answers as well, however you made it complicated by using both methods of solving simultaneous equations. Stick to one method to keep it simple.

**HallsofIvy** · December 14th 2010, 03:30 AM

I don't know what you mean by "Substitute the second line in the first one then solve the system". You don't substitute "lines" or "equations" you substitute for specific quantities.

Since there is now no $x^2$ in the first equation, I would solve for $x= -4/3(1+ y^2)$ and substitute that for x in the second equation:
$(16/9)(1+ 2y^2+ y^4)- y^2= 4$
$16+ 32y^2+ 16y^4- 9y^2= 36$
$16y^4+ 23y^2- 20 = 0$

That's now a quadratic equation for $y^2$ . One of the two roots will be negative which is impossible since y is real. The positive root will give two roots for y.

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