Solve the system algebraically:
32x^2 + 9y^2 = 324
3y^2 - x^2 = 3

Question

Solve the system algebraically:
32x^2 + 9y^2 = 324
3y^2 - x^2 = 3

displayerror · Answer

You can solve by substitution. Either solve for x in terms of y or y in terms of x using one equation, then substitute that into the other equation to solve.

Using the second equation, if we solve for x first:
$$3y^2-x^2=3$$
$$x^2 = 3y^2-3$$
Now we have x^2 in terms of y^2. Notice that there is also an x^2 term in the first equation. We can then substitute the expression for x^2 into the first equation wherever we see x^2:
The first equation:
$$32x^2+9y^2=324$$
Substituting in for x^2:
$$32\left( 3y^2-3 \right) + 9y^2 = 324$$
Distribute the 32:
$$96y^2 - 96 + 9y^2 = 324$$
Combine like terms to solve for y. Once we find y, we can plug that into our expression for x^2 to solve for x
$$x^2 = 3y^2 -3$$

anonymous · Answer

@DisplayError I got $$y = \pm4  $$    and    $$y = \pm6.7$$
Is that what I was supposed to get or did you get something different?

anonymous · Answer

That second equation is supposed to be $$x = \pm6.7$$

displayerror · Answer

You're close! Remember, we need to solve for y, not y^2. 
From this equation:
$$96y^2 - 96 + 9y^2 = 324$$
We can combine the y^2 terms and move the 96 to the other side:
$$105y^2 = 420$$
Now solve for y, not y^2.

anonymous · Answer

Oh it would be y = +- 2!

displayerror · Answer

Yep! Now plug into the equation for x^2 to solve for x:
$$x^2 = 3y^2 -3$$

anonymous · Answer

And x would be $$x=\pm3$$  so does that mean it intersects at 4 points?

displayerror · Answer

Yep! You should get the points (2, 3), (2, -3), (-2, 3), and (-2, -3), so four points of intersection.

displayerror · Answer

I've attached a plot of the two polynomials to help you see how it looks like