SOLUTION: I tried to follow the directions on the video, but I got lost somewhere and messed up. I am lost. Solve each of the following systems. (1) x + y = 5 x^2 + y = 5 (2) x

Question 62040: I tried to follow the directions on the video, but I got lost somewhere and messed up. I am lost.
Solve each of the following systems.
(1) x + y = 5
x^2 + y = 5
(2) x - y = 2
x^2 + y^2 = 4
(3) x^2 + y^2 = 12
x^2 + y = 10
(4) x^3 - 2^x - 5X + 6 = 0
3x - y + 6 = 0
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
(1)
x + y = 5 is also ~> y = 5 - x
x^2 + y = 5
x^2 + 5 - x = 5 Remember you can substitute the 5 - x in for y
x^2 - x = 0
x(x - 1) = 0
x = 0 and x = 1
Plug:
y = 5 - x
(0,5) and (1,4)

(2)
x - y = 2 is also ~> x = 2 + y
x^2 + y^2 = 4
(2 + y)^2 + y^2 = 4
4 + 4y + y^2 + y^2 = 4
2y^2 + 4y = 0
y^2 + 2y = 0
y(y + 2) = 0
y = 0 and y = -2
Plug:
x = 2 + y
(2,0) and (0,-2)

(3)
x^2 + y^2 = 12 is also ~> x^2 = 12 - y^2
x^2 + y = 10
12 - y^2 + y = 10
-y^2 + y + 2 = 0
(y + 1)(-y + 2) = 0
y = -1 and y = 2
Plug:
x^2 = 12 - y^2
x = +- sqrt(12 - y^2)
x = +- sqrt(12 - 1) = +- sqrt(11)
(sqrt(11),-1)
(-sqrt(11),-1)
x = +- sqrt(12 - y^2) = +- sqrt(12 - 4) = +- sqrt(8)
(sqrt(8),2)
(-sqrt(8),2)

(4)
3x - y + 6 = 0 ~> red line demonstrates this line
x^3 - x^2 - 5x + 6 = 0 ~> found from green line (x-intercepts are answers)

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