SOLUTION: Solve the system of equations for real values of x and y. (Sort your answers in increasing order, first by their x -values, then by their y -values.) 2x^2-y^2+7= 0 3x^2-4y^2+3

Algebra ->  Equations -> SOLUTION: Solve the system of equations for real values of x and y. (Sort your answers in increasing order, first by their x -values, then by their y -values.) 2x^2-y^2+7= 0 3x^2-4y^2+3      Log On


   

Question 329514: Solve the system of equations for real values of x and y. (Sort your answers in increasing order, first by their x -values, then by their y -values.)
2x^2-y^2+7= 0
3x^2-4y^2+33=0
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.2x%5E2-y%5E2%2B7=+0
2.3x%5E2-4y%5E2%2B33=0
From eq. 1,
y%5E2=2x%5E2%2B7
4y%5E2=8x%5E2%2B28
Substitute into eq. 2,
3x%5E2-8x%5E2-28%2B33=0
-5x%5E2%2B5=0
x%5E2=1
x=0+%2B-+1
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For each x, find the y value.
y%5E2=2x%5E2%2B7
y%5E2=2%2B7
y=0+%2B-+3
(1,3),(1,-3),(-1,3),(-1,-3)
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