1. ## solve the equation

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.

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

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

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

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

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

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

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

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.

3. 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 $\displaystyle x^2$ in the first equation, I would solve for $\displaystyle x= -4/3(1+ y^2)$ and substitute that for x in the second equation:
$\displaystyle (16/9)(1+ 2y^2+ y^4)- y^2= 4$
$\displaystyle 16+ 32y^2+ 16y^4- 9y^2= 36$
$\displaystyle 16y^4+ 23y^2- 20 = 0$

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