Question

Use the substitution method to find the solutions to the linear-quadratic system y = x^2 – x – 40, y = -3x – 5?

A. (5, -10)
B. (5, -10) and (-7, 26)
C. (5, -20) and (-7, 16)
D. (-7, 26)

Answer 1

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

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

Since both equations are equal to \( y \), we can set them equal to each other:

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

Now, we'll rearrange the equation:

1. Move all terms to one side of the equation:

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

This simplifies to:

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

2. Next, we factor the quadratic equation:

\[
(x + 7)(x - 5) = 0
\]

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

\[
x + 7 = 0 \quad \Rightarrow \quad x = -7
\]

\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]

3. Now we can find the corresponding values of \( y \) for both values of \( x \):

a. For \( x = 5 \):

Using Equation 2:

\[
y = -3(5) - 5 = -15 - 5 = -20
\]

So one solution is \( (5, -20) \).

b. For \( x = -7 \):

Again using Equation 2:

\[
y = -3(-7) - 5 = 21 - 5 = 16
\]

So the other solution is \( (-7, 16) \).

4. The solutions to the system are:

- \( (5, -20) \)
- \( (-7, 16) \)

Therefore, the correct answer is **C. (5, -20) and (-7, 16)**.