What are the x-coordinates of the solutions to this system of equations? x2 + y2 = 16 y = x + 4

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What are the x-coordinates of the solutions to this system of equations? x2 + y2 = 16 y = x + 4

Mathematics
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do a substitution y = x + 4 means we can replace "y" with "x + 4" when we see y, so x^2 + y^2 = 16 x^2 + (x+4)^2 = 16 know where to go from here?
then u divide the by 4
uhhhh not quite

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Other answers:

then we can use FOIL to expand (x+4)^2
2x+8
not quite, the ^2 is an exponent (x+4)^2 = (x+4)(x+4) = ?
0
nope, not quite... hint: (x+A)(x+A) = x^2 + 2*A*x + A^2 (x+4)(x+4) = ____________________ fill in the blank
x+4x+4x+16
almost, you're very close! x^2 + 4x + 4x + 16
that simplifies to: x^2 + 8x + 16, with me so far?
2x+8x+16 4x+16
nope, not quite. please remember that x^2 is NOT equal to 2x
x^2 means x*x, ok?
so it was x+8x+16
confuse
yes, x^2 + 8x + 16, with me so far?
x+4^2
no, we just leave it as x^2 + 8x + 16, ok?
What are the x-coordinates of the solutions to this system of equations? x2 + y2 = 16 y = x + 4 −4 and 0 4 and 0 −2 and −3 2 and −3
those r the option
right, we'll get to that in a minute, ok?
right now we have: (x+4)^2 = x^2 + 8x + 16 plugging that back into the first equation gives us x^2 + x^2 + 8x + 16 = 16 know where to go from here?
no not really
start by subtracting 16 from each side.
0
what about the left side? x^2 + x^2 + 8x + 16 - 16 = ?
4
nope, maybe this will make things easier: (x^2 + x^2 + 8x) + 16 - 16 = ?
8x^2+0
2x^2+8x
good, 2x^2 + 8x
now we have 2x^2 + 8x = 0 solve for x
4x+8x=0 4x=0
remember what I said earlier? x^2 is NOT 2x
2x^2 + 8x = 0 2x(x+4) = 0 solve for x
2x2+8x
not quite... 2x^2 + 8x is what I gave you to start with
2x = 0 x = ?
4
try again
2x = 0 divide both sides by 2
0
good
now we solve (x+4) = 0 x = ?
4
almost but not quite
subtract 4 from each side
-4
good, so our solutions are 0 and -4
thank u so much

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