System of equations, help

x^2+y^2=24
y^2-3x^2=12

I need help solving this

Solve the top equation for y^2: y^2=24-x^2

Then substitute into bottom equation: 24-4x^2=12.

Solve for x: x^2=3 , x=√3 or -√3

Plug back into either equation to get y: y^2=21 , y=√21 or -√21

well, there is 3 ways to solve for a system of equation:
1) Graph
2)Substitution
3) Elimination

My favorite is elimination because it involves the least amount of steps. Elimination means to eliminate one of the two variables, and then you can solve for the other variable. For example:

x^2+y^2=24
- y^2 + 3x^2= 12
--------------------------------------
4x^2= 12 (in the step above I subtracted the two equations)
-- ---
4 4
x^2=3
x= radical 3

Now, I'll use substitution. I'll substitute x for radical 3, then solve for y.

(radical 3)^2+y^2=24
3+y^2=24
-3 -3
--------------
y^2= 21
y= radical 21

Your answer is (radical 3, radical 7)

By doing eqn1-eqn2, you can get rid of y.
(x^2+y^2=24)
-(-3x^2+y^2=12)
=(4x^2 =12)

the rest is easy
4x^2=12
x^2=3

x^2+y^2=24
3+y^2=24
y^2=21