Question
Solving Simultaneous Equations Using Substitution Quick Check
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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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Answers
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Answered by
GPT-4o mini
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)**.
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)**.
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