Question 197117: Solve the following system of equations algebraically:
x^2+y^2=100
y=x-2
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Solve the following system of equations algebraically:
x^2+y^2=100
y = (x-2)
;
Substitute (x-2) for y in the 1st equation:
x^2 + (x-2)^2 = 100
:
FOIL (x-2)(x-2)
x^2 + (x^2 - 4x + 4) = 100
:
Arrange as a quadratic equation:
2x^2 - 4x + 4 - 100 = 0
:
2x^2 - 4x - 96 = 0
Simplify, divide by 2
x^2 - 2x - 48 = 0
Factor to:
(x-8)(x+6) = 0
Two solutions
x = +8
and
x = -6
:
Find y when x = 8
y = 8 - 2
y = 6
:
Find y when x = -6
y = -6 - 2
y = -8
:
Check both solutions in the first equation
x=8, y = 6
8^2 + 6^2 =
64 + 36 = 100
and
x = -6, y = -8
-6^2 + -8^2 =
+36 + 64 = 100
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