SOLUTION: I have a system of nonlinear equations, x^2+y^2=49 y^2-3x=49 I need to find the real solutions... Please help!

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Question 1050549: I have a system of nonlinear equations,
x^2+y^2=49
y^2-3x=49
I need to find the real solutions... Please help!
Found 2 solutions by advanced_Learner, jorel555:
Answer by advanced_Learner(501)   (Show Source): You can put this solution on YOUR website!
I have a system of nonlinear equations,
x^2+y^2=49
y^2-3x=49
I need to find the real solutions... Please help!
---------------------------------------------------
it is straighforward
49-x^2-3x=49
x^2+3x=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=9 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0, -3. Here's your graph:
Answer by jorel555(1290)   (Show Source): You can put this solution on YOUR website!
x²+y²=49
y²-3x=49;
Subtracting the second equation from the first, we get:
x²+3x=0
x²=-3x
x=-3
49-9=y²
y²=40
y=√40=2√10. ☺☺☺☺
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