What is the solution of the linear-quadratic system of equations? Show your work
{y = x^2 + 2x - 3
{y = 2x + 1
To find the solution of the linear-quadratic system of equations, we need to set the two equations equal to each other and solve for x first.
So we have:
x^2 + 2x - 3 = 2x + 1
To simplify, we can subtract 2x and 1 from both sides of the equation:
x^2 + 2x - 2x - 3 - 1 = 0
Simplifying further:
x^2 - 4 = 0
Now we can factor the quadratic equation:
(x + 2)(x - 2) = 0
Setting each factor equal to zero gives us two possible values for x:
x + 2 = 0
x = -2
x - 2 = 0
x = 2
Therefore, the solutions for x are x = -2 and x = 2.
To find the corresponding y-values, we can substitute these values of x back into either of the original equations. Let's use the first equation y = x^2 + 2x - 3:
For x = -2:
y = (-2)^2 + 2(-2) - 3
y = 4 - 4 - 3
y = -3
For x = 2:
y = (2)^2 + 2(2) - 3
y = 4 + 4 - 3
y = 5
Therefore, the solutions to the linear-quadratic system of equations are (-2, -3) and (2, 5).