SOLVE THE NONLINEAR SYSTEM OF EQUATIONS FOR REAL SOLUTIONS

y = x

^{2}+ 3y = -x

^{2}+ 5SOLVE THE NONLINEAR SYSTEM OF EQUATIONS FOR REAL SOLUTIONS

y = x^{2} + 3

y = -x^{2} + 5

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Houston, TX

Hi, Mario!

Solve the system of equations using substitution. Since both equations are already in the "y=" form, we can set x^{2}+3 equal to -x^{2}+5:

x^{2 }+ 3 = -x^{2} + 5

Rearrange the terms:

2x^{2} = 2

Solving, we get that x = 1 or x = -1. But we are not done yet! Solving the system means getting x and y values.

We can plug the values we found for x into either of the original equations. I recommend trying both equations just to make sure that your answers do work in both.

We get y=4 for both x=1 and x=-1 in both equations. So
the solution is (1,4) and (-1,4).

Hope this helps!

Kathye P.

Ansonia, CT

You can set the two expressions for y equal to each other and get

x^{2} + 3 = -x^{2} +5 which simplifies to

2x^{2} = 5 - 3 = 2 so

x^{2} = 1 and

x = ±1

plugging this into either of the equations yields y = 4

so these two quadratics intersect at (±1, 4)

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