Solve the linear-quadratic system of equations algebraically. Show your step-by-step work.
y=−x^2+2x−3
y=4x−3
To solve the linear-quadratic system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously.
First, we can set the two equations equal to each other since they both equal y:
−x^2+2x−3 = 4x−3
Next, we can simplify the equation:
−x^2+2x−3 = 4x−3
-x^2 + 2x = 4x
Moving all the terms to one side of the equation:
-x^2 - 4x + 2x = 0
Combining like terms:
-x^2 - 2x = 0
Factoring out an x:
x(-x - 2) = 0
Setting each factor equal to zero:
x = 0 or -x - 2 = 0
If x = 0:
Using one of the original equations, y = 4x - 3:
y = 4(0) - 3
y = -3
If -x - 2 = 0:
-x = 2
x = -2
Using one of the original equations, y = 4x - 3:
y = 4(-2) - 3
y = -11
Therefore, the solutions to the linear-quadratic system of equations are (0, -3) and (-2, -11).