Solving Simultaneous Equations Using Substitution Quick Check

Question

5 of 55 of 5 Items
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,−10) and (−7,26)
left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis

(−7,26)
left parenthesis negative 7 comma 26 right parenthesis

(5,−10)
left parenthesis 5 comma negative 10 right parenthesis

(5,−20) and (−7,16)
left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis
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Answer 1

To solve the system of equations using substitution, we start with the two equations given:

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

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

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

Next, we will move all terms to one side to form a quadratic equation:

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

This simplifies to:

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

Now we factor the quadratic equation:

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

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

Next, we substitute these \( x \)-values back into one of the original equations to find the corresponding \( y \)-values. We will use the second equation \( y = -3x - 5 \):

1. For \( x = -7 \): \[ y = -3(-7) - 5 = 21 - 5 = 16 \] So one solution is \( (-7, 16) \).

2. For \( x = 5 \): \[ y = -3(5) - 5 = -15 - 5 = -20 \] So the other solution is \( (5, -20) \).

Thus, the solutions to the system of equations are:

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

Looking at the provided answer choices:

• (5,−20) and (−7,16) matches our solutions.

Therefore, the correct response is:

(5,−20) and (−7,16).