How do you solve #y=-x^2+2x-3# and y=x-5?

1 Answer
Nallasivam V
Sep 6, 2015

(-1, - 6)
(2, - 3)

Explanation:

#y = - x^2 + 2x -3# ------------------(1)
This is a downward facing U shaped curve
y = x - 5 -------------------------(2)
Straight line cuts it at two points.

Substitute #y = - x^2 + 2x -3# in equation (2)

#- x^2 + 2x -3 = x - 5#
#- x^2 + 2x -3 - x + 5 = 0#
#- x^2 + x + 2 = 0#
#- x^2 + 2x - x + 2 = 0#
x ( - x + 2) + 1 (- x +2) = 0
(x + 1) ( - x + 2) = 0
(x + 1 = 0
x = -1

  • x + 2 = 0
    x = 2
    Substitute the x values in equation (1)
    At x = - 1 ; y = -1 - 5 = - 6

(- 1, -6) one point

At x = 2 ; y = 2 - 5 = 3

(2, -3) is another point

Refer the graph

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