Asked by Hana blanie
Which of the following uses the substitution method to solve linear-quadratic system y=x2+3x+5, y=2x+11?(1 point)
Responses
(−3,5)
left parenthesis negative 3 comma 5 right parenthesis
(−2,7) and (3,17)
left parenthesis negative 2 comma 7 right parenthesis and left parenthesis 3 comma 17 right parenthesis
(2,15) and (−3,5)
left parenthesis 2 comma 15 right parenthesis and left parenthesis negative 3 comma 5 right parenthesis
(2,15)
Responses
(−3,5)
left parenthesis negative 3 comma 5 right parenthesis
(−2,7) and (3,17)
left parenthesis negative 2 comma 7 right parenthesis and left parenthesis 3 comma 17 right parenthesis
(2,15) and (−3,5)
left parenthesis 2 comma 15 right parenthesis and left parenthesis negative 3 comma 5 right parenthesis
(2,15)
Answers
Answered by
GPT-4o mini
To solve the linear-quadratic system using the substitution method, we can start with the equations you've provided:
1. \( y = x^2 + 3x + 5 \)
2. \( y = 2x + 11 \)
Since both equations equal \( y \), we can set them equal to each other:
\[
x^2 + 3x + 5 = 2x + 11
\]
Next, we can rearrange this equation to form a standard quadratic equation:
\[
x^2 + 3x - 2x + 5 - 11 = 0
\]
This simplifies to:
\[
x^2 + x - 6 = 0
\]
Now we can factor the quadratic equation:
\[
(x + 3)(x - 2) = 0
\]
Setting each factor to zero gives us the solutions for \( x \):
1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 2 = 0 \) → \( x = 2 \)
Now we will find the corresponding \( y \) values for each \( x \).
1. For \( x = -3 \):
\[
y = 2(-3) + 11 = -6 + 11 = 5
\]
This gives us the point \( (-3, 5) \).
2. For \( x = 2 \):
\[
y = 2(2) + 11 = 4 + 11 = 15
\]
This gives us the point \( (2, 15) \).
Thus, the solutions to the system are \( (-3, 5) \) and \( (2, 15) \).
From the options given, the correct response is:
**(2, 15) and (−3, 5)**.
1. \( y = x^2 + 3x + 5 \)
2. \( y = 2x + 11 \)
Since both equations equal \( y \), we can set them equal to each other:
\[
x^2 + 3x + 5 = 2x + 11
\]
Next, we can rearrange this equation to form a standard quadratic equation:
\[
x^2 + 3x - 2x + 5 - 11 = 0
\]
This simplifies to:
\[
x^2 + x - 6 = 0
\]
Now we can factor the quadratic equation:
\[
(x + 3)(x - 2) = 0
\]
Setting each factor to zero gives us the solutions for \( x \):
1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 2 = 0 \) → \( x = 2 \)
Now we will find the corresponding \( y \) values for each \( x \).
1. For \( x = -3 \):
\[
y = 2(-3) + 11 = -6 + 11 = 5
\]
This gives us the point \( (-3, 5) \).
2. For \( x = 2 \):
\[
y = 2(2) + 11 = 4 + 11 = 15
\]
This gives us the point \( (2, 15) \).
Thus, the solutions to the system are \( (-3, 5) \) and \( (2, 15) \).
From the options given, the correct response is:
**(2, 15) and (−3, 5)**.
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