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.
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)
To solve the system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously.
Let's start by setting the two equations equal to each other:
-x^2 + 4x - 3 = -x + 1
Now, we can rearrange the equation to form a quadratic equation:
-x^2 + 4x - x - 3 - 1 = 0
-x^2 + 3x - 4 = 0
Next, we need to solve this quadratic equation. There are several ways to do this, but let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -1, b = 3, and c = -4. Substituting these values into the quadratic formula, we get:
x = (-(3) ± √((3)^2 - 4(-1)(-4))) / (2(-1))
x = (-3 ± √(9 - 16)) / -2
x = (-3 ± √(-7)) / -2
Since we have a negative value inside the square root, the quadratic equation does not have real solutions. Therefore, the system of equations has no real solutions.
It’s poop
You poop and u get 69
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)