What is the solution of the linear-quadratic system of equations?{y=x^2+5x−3y−x=2

Answer 1

To find the solution of the linear-quadratic system of equations, we'll solve the two equations simultaneously.

1) Rearrange the first equation:
y = x^2 + 5x - 3

2) Substitute the rearranged first equation into the second equation:
x^2 + 5x - 3 - x = 2

3) Simplify the equation:
x^2 + 4x - 5 = 0

4) Factor the equation:
(x + 5)(x - 1) = 0

5) Set each factor equal to zero and solve for x:
x + 5 = 0 or x - 1 = 0
x = -5 or x = 1

6) Substitute the values of x back into the first equation to find the corresponding y values:
For x = -5:
y = (-5)^2 + 5(-5) - 3
y = 25 - 25 - 3
y = -3

For x = 1:
y = 1^2 + 5(1) - 3
y = 1 + 5 - 3
y = 3

Therefore, the solution to the linear-quadratic system of equations is:
x = -5, y = -3
x = 1, y = 3