Question 467084: Solve the following system for real number solutions using elimination or substitution. I am just trying to check my work. Here is what I got:
x^2+y^2=9
2x-y=3
Answer:
using substitution y=2x-3
x^2+(2x-3)^2=9
x^2+4x^2-12x+9=9
5x^2-12x=0
x(5x-12)=0
x=0 and x=12/5
Therefore:
(0,-3) and (12/5,9/5)
But I feel like I am missing something. When I graph on my calculator there is only one intersection but both answers seem to work. Plus, I am concerned about the ^2, did I capture all possible real number answers?
Thanks!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! x^2+y^2=9
2x-y=3
using substitution y=2x-3
x^2+(2x-3)^2=9
x^2+4x^2-12x+9=9
5x^2-12x=0
x(5x-12)=0
x=0 and x=12/5
Therefore:
(0,-3) and (12/5,9/5)
But I feel like I am missing something. When I graph on my calculator there is only one intersection but both answers seem to work. Plus, I am concerned about the ^2, did I capture all possible real number answers?
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It's a straight line and a circle, so 2 intersections are the max #.
It's complete and correct.
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