Question

What are the x-coordinates of the solutions to this system of equations?

x2 + y2 = 16
y = x + 4

Answer 1

y=x+4
sub x+4 for y in other equation
x²+(x+4)²=16
expand
x²+x²+8x+16=16
2x²+8x+16=16
minus 16 both sides
2x²+8x=0
factor
2x(x+4)=0
set each to zero
2x=0
x=0

x+4=0
x=-4


x=0 or -4

the x coordiantes woud be at x=0 and x=-4

Answer 2

What this question is asking is at what points do these function/relation intersect. You can solve this by graphing it, or solve it algebraically.

You are given x^2 + y^2 = 16 and y = x + 4

You can substitute y = x + 4 into x^2 + y^2 = 16
So we get

x^2 + (x+4)^2 = 16

This simplifies into

x^2 + x^2 + 8x + 16 = 16

Combine like terms

2x^2 + 8x = 0

Then factor 2x out:

2x( x + 4 ) = 0

We can see that our solution is x = 0, x = -4

So our answer is

x = 0, x = -4
Hopes this helps!