Solve the followin system algebrically.
y= 1/2(x-3)2+4
y= -1/2(x-1)2+6
6 years ago
Multiply both equations by 2 to eliminate the fraction:
2y = (x-3)^2 + 8
2y =-(x-1)^2 + 12
since both equations equal 2y we can set them equal to each other and solve.
(X-3)^2 + 8 = -(x-1)^2 + 12
x^2 -6x+9+8 = -(x^2 - 2x +1)+12
x^2 -6x+17 = -x^2 + 2x -1 +12
x^2 -6x+17.= -x^2 +2x +11
2x^2 -8x + 6 = 0
x^2 - 4x + 3 = 0
(x-3)(x-1) =0
x= 3 and x =1
6 years ago
Answered By Jesslyn C
Multiply both equations by 2 to eliminate the fraction:
2y = (x-3)^2 + 8
2y =-(x-1)^2 + 12
since both equations equal 2y we can set them equal to each other and solve.
(X-3)^2 + 8 = -(x-1)^2 + 12
x^2 -6x+9+8 = -(x^2 - 2x +1)+12
x^2 -6x+17 = -x^2 + 2x -1 +12
x^2 -6x+17.= -x^2 +2x +11
2x^2 -8x + 6 = 0
x^2 - 4x + 3 = 0
(x-3)(x-1) =0
x= 3 and x =1