Questions on Algebra: Linear Equations, Graphs, Slope answered by real tutors!

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Question 170266: Help On This Will Be Great Appreciated<3
Solve The Non-Linear system of Equations. Give Exact Values.

x^2+y^2=10
3y=x^2+8
: Help On This Will Be Great Appreciated<3
Solve The Non-Linear system of Equations. Give Exact Values.

x^2+y^2=10
3y=x^2+8

Answer by Earlsdon(3816) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
x^2+y^2 = 10
3y = x^2+8 Rewrite this as x^2 = 3y-8 and substitute into the 1st equation.
(3y-8)+y^2 = 10 Simplify.
y^2+3y-18 = 0 Factor.
(y-3)(y+6) = 0 Apply the zero product rule.
y = 3 or y = -6
Use the positive solution and substitute into the second equation to solve for x.
3y = x^2+8 Substitute y = 3.
3(3) = x^2+8 Subtract 8 from both sides.
9-8 = x^2 Take the square root of both sides.
x = 1 or x = -1
The solution consists of two points of intersection:
(1, 3) and (-1, 3)
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Just for interest, since we had two solutions for y (y = 3 and y = -6), let's see what happens when we choose the second solution of y = -6 and substitute into the first equation to solve for x.
3y = x^2+8 Substitute y = -6.
3(-6) = x^2+8 Subtract 8 from both sides.
x^2 = -26 Take the square root of both sides.
x = sqrt(-26) or x = -sqrt(-26) and, as you can see, the results are "imaginary" numbers, so there is no "real" solution for y = -6.
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Let's see what the graph of these two functions look like:
graph(400,400,-5,5,-5,5,sqrt(-x^2+10),-sqrt(-x^2+10),(1/3)x^2+8/3)