SOLUTION: Please help me to find the solution set(s) to this problem. Thank you! {{{3x^2+2y^2-29=0}}} {{{2x^2-5y^2-13=0}}}

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Please help me to find the solution set(s) to this problem. Thank you! {{{3x^2+2y^2-29=0}}} {{{2x^2-5y^2-13=0}}}      Log On


   

Question 430768: Please help me to find the solution set(s) to this problem. Thank you!
3x%5E2%2B2y%5E2-29=0
2x%5E2-5y%5E2-13=0
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B2y%5E2-29=0
2x%5E2-5y%5E2-13=0
Do it by the addition method.
Eliminate y%5E2 by multiplying the first equation through by 5
and the second equation through by 2.


red%285%293x%5E2%2Bred%285%292y%5E2-red%285%2929=red%285%290
red%282%292x%5E2-red%282%295y%5E2-red%282%2913=red%285%290


15x%5E2%2B10y%5E2-145=0
 4x%5E2-10y%5E2-26=0
19x%5E2-171=0
x%5E2-9=0
x%5E2=9
x=%22%22%2B-+3 


Substitute ±3 for x in

3x%5E2%2B2y%5E2-29=0
3%28%22%22%2B-+3%29%5E2%2B2y%5E2-29=0
3%289%29%2B2y%5E2-29=0
27%2B2y%5E2-29=0
2y%5E2-2=0
y%5E2-1=0
y%5E2=1
y=%22%22%2B-+1

So there are 4 solutions:

(3,1), (3,-1), (-3,1), and (-3,-1)

Here is the graphical solution:



The black oval curve is the graph of the first equation.
The red curve (in two parts) is the graph of the second equation
Notice that they intersect at those 4 solution points.

Edwin