Question

Determine the solution to the linear-quadratic system. "y" equals "x" squared plus 2 "x" plus negative 5y = x2+2x+−5 "y" equals negative 2 "x" minus 5y =−2​x−5 The solution(s) are: (Select all that apply.)

Answer 1

To solve the given system of equations, we will first rewrite the equations clearly:

1. \( y = x^2 + 2x - 5 \) (Equation 1)
2. \( y = -2x - 5 \) (Equation 2)

Next, substitute Equation 2 into Equation 1:

\[
-2x - 5 = x^2 + 2x - 5
\]

Now we can simplify the equation:

1. Add 5 to both sides:
\[
-2x = x^2 + 2x
\]

2. Rearranging the equation:
\[
0 = x^2 + 2x + 2x
\]
\[
x^2 + 4x = 0
\]

3. Factoring:
\[
x(x + 4) = 0
\]

This gives us the solutions for \(x\):

- \(x = 0\)
- \(x = -4\)

Now, we substitute these \(x\) values back into Equation 2 to find the corresponding \(y\) values.

1. For \(x = 0\):
\[
y = -2(0) - 5 = -5
\]
So, one solution is \((0, -5)\).

2. For \(x = -4\):
\[
y = -2(-4) - 5 = 8 - 5 = 3
\]
So, the other solution is \((-4, 3)\).

Thus, the solutions to the linear-quadratic system are:

- \((0, -5)\)
- \((-4, 3)\)

**Final answer:**
The solutions are \((0, -5)\) and \((-4, 3)\).