Question

9. Solve the system of equations algebraically. Show all work. (2 pts.)
y = -x2 + 4x – 3
y = -x + 1
I have tried everything before coming here, but I'm still stuck.

Answer 1

since y = -x + 1 and y = -x^2 + 4x – 3 are both equal to y, you can set them opposite to each other because they are both equal

you then get
-x+1 = -x2 + 4x – 3

if you multiply both sides by -1, you get
x-1 = x^2 - 4x + 3

after moving everything to one side, you get
0 = x^2 - 5x + 4

after that, it's just a matter of factoring
0 = (x-1)(x-4)

we get x = 1, and x = 4

If you plug both of these x values into the equation, you come up with y = 0 for x = 1, and y = -3 for x = 4

you answers are (0,1) and (-3,4)

Answer 2

It’s poop

Answer 3

You poop and u get 69

Answer 4

x² - 4x + 3 = x -1
-x. + 1
x² - 5x + 4 = 0
(x - 4)(x-1) = 0
x - 4 = 0
+4. +4
x = 4
x - 1 = 0
+1. +1
x = 1
(1²) -4(1) + 3 = 0
(4²) - 4(4) + 3 =3
(x₁︎,y₁︎)(x₂︎,y₂︎) = (1, 0) (4,3)