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Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40 , y=−3x−5 ?(1 point) Responses (5,−1...Question
Use the substitution method to find the solutions to the linear-quadratic system y=x^2−x−40, y=−3x−5?(1 point)
Responses
(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
(5,−10) and (−7,26)
left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis
(−7,26)
Responses
(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
(5,−10) and (−7,26)
left parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis
(−7,26)
Answers
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Answered by
GPT-4o mini
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)
Step 1: Set the two equations equal to each other:
\[
x^2 - x - 40 = -3x - 5
\]
Step 2: Rearrange the equation to set it to zero:
\[
x^2 - x + 3x - 40 + 5 = 0
\]
\[
x^2 + 2x - 35 = 0
\]
Step 3: Factor the quadratic equation:
\[
(x + 7)(x - 5) = 0
\]
Step 4: Set each factor equal to zero and solve for \( x \):
1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)
Step 5: Now substitute each value of \( x \) back into one of the original equations to find the corresponding \( y \) values. We'll use equation 2 (\( y = -3x - 5 \)).
For \( x = -7 \):
\[
y = -3(-7) - 5 = 21 - 5 = 16
\]
So, one solution is \( (-7, 16) \).
For \( x = 5 \):
\[
y = -3(5) - 5 = -15 - 5 = -20
\]
So, another solution is \( (5, -20) \).
Thus, the solutions to the system are:
1. \( (5, -20) \)
2. \( (-7, 16) \)
Thus, the correct response is:
**(5,−20) and (−7,16)**.
1. \( y = x^2 - x - 40 \) (equation 1)
2. \( y = -3x - 5 \) (equation 2)
Step 1: Set the two equations equal to each other:
\[
x^2 - x - 40 = -3x - 5
\]
Step 2: Rearrange the equation to set it to zero:
\[
x^2 - x + 3x - 40 + 5 = 0
\]
\[
x^2 + 2x - 35 = 0
\]
Step 3: Factor the quadratic equation:
\[
(x + 7)(x - 5) = 0
\]
Step 4: Set each factor equal to zero and solve for \( x \):
1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)
Step 5: Now substitute each value of \( x \) back into one of the original equations to find the corresponding \( y \) values. We'll use equation 2 (\( y = -3x - 5 \)).
For \( x = -7 \):
\[
y = -3(-7) - 5 = 21 - 5 = 16
\]
So, one solution is \( (-7, 16) \).
For \( x = 5 \):
\[
y = -3(5) - 5 = -15 - 5 = -20
\]
So, another solution is \( (5, -20) \).
Thus, the solutions to the system are:
1. \( (5, -20) \)
2. \( (-7, 16) \)
Thus, the correct response is:
**(5,−20) and (−7,16)**.
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