Question

Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point) Responses (5,−20) and (−7,16) left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis (−7,26) left parenthesis negative 7 comma 26 right parenthesis (5,−10) and (−7,26) left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis (5,−10)

Answer 1

To solve the system of equations using the substitution method, we have the following two equations:

1. \( y = x^2 - x - 40 \) (Equation 1)
2. \( y = -3x - 5 \) (Equation 2)

We will substitute Equation 2 into Equation 1.

**Step 1: Substitute \( y \) from Equation 2 into Equation 1.**

So, we take Equation 1:

\[
-3x - 5 = x^2 - x - 40
\]

**Step 2: Rearrange the equation.**

Move all terms to one side:

\[
x^2 - x + 3x + 40 - 5 = 0
\]

Combine like terms:

\[
x^2 + 2x + 35 = 0
\]

**Step 3: Solve for \( x \) using the quadratic formula.**

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \( a = 1 \), \( b = 2 \), and \( c = 35 \):

First, calculate the discriminant:

\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 35 = 4 - 140 = -136
\]

Since the discriminant is negative, there are no real solutions for \( x \).

**Step 4: Conclusion**

Since the quadratic equation does not yield any real solutions for \( x \), the linear-quadratic system has no intersection points that result in real coordinates \( (x,y) \).

Therefore, the answer is that there are no solutions to the system of equations. If one of the pairs of responses you provided is meant to represent complex solutions, please clarify, as the provided options seem to suggest otherwise.