Question 63370
How do I solve this problem using system of equations?

E1:  x^2 + y^2 = 25
E2:  y^2 + x = 5
:
Solve E2 for y^2 and substitute that into E1 and solve for x.
E2:  y^2+x-x=5-x ---> y^2=5-x
:
E1:  x^2+(5-x)=25
Solve E1 like you would any quadratic equation.
x^2-x+5=25
x^2-x+5-25=25-25
x^2-x-20=0
(x-5)(x+4)=0
x-5=0 and x+4=0
x-5+5=0+5 and x+4-4=0-4
x=5 and x=-4
:
Substitute those back into E2 and solve for y.
y^2+(5)=5
y^2+5-5=5-5
y^2=0
{{{sqrt(y^2)=+-sqrt(0)}}}
y=0
One solution is (5,0)
y^2+(-4)=5
y^2-4=5
y^2-4+4=5+4
y^2=9
{{{sqrt(y^2)=+-sqrt(9)}}}
y=+\-3
These two solutions are (-4,-3) and (-4,3)
So there are three points of intersection:
(5,0), (-4,-3), and (-4,3)
I tried to show you graphically, but the program is not cooperating.  If you have a graphing calculator, graph:
y1:sqrt(25-x^2)
y2:-sqrt(25-x^2)
y3:sqrt(5-x)
y4: -sqrt(5-x)
Set your window to x-min:-15,x-max: 15, y-min: -10, y-max: 10
You'll see a circle (E1) intersected by a sideways parabola (E2) in three places:(5,0), (-4,-3), and (-4,3)
Happy Calculating!!!