Question
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Part 1
Solve the system of equations.
left brace Start 2 By 1 Matrix 1st Row 1st Column y equals x squared minus 7 x plus 3 2nd Row 1st Column y equals negative x minus 6 EndMatrix y=x2−7x+3 y=−x−6
Question content area bottom
Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is(are) enter your response here.
(Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.)
Part 1
Solve the system of equations.
left brace Start 2 By 1 Matrix 1st Row 1st Column y equals x squared minus 7 x plus 3 2nd Row 1st Column y equals negative x minus 6 EndMatrix y=x2−7x+3 y=−x−6
Question content area bottom
Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is(are) enter your response here.
(Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.)
Answers
Answered by
GPT-4o mini
To solve the system of equations given:
1. \( y = x^2 - 7x + 3 \)
2. \( y = -x - 6 \)
We can set the two equations equal to each other since they both equal \( y \):
\[
x^2 - 7x + 3 = -x - 6
\]
Now, let's rearrange this equation to form a standard quadratic equation. First, move all terms to one side:
\[
x^2 - 7x + 3 + x + 6 = 0
\]
This simplifies to:
\[
x^2 - 6x + 9 = 0
\]
Next, we can factor this quadratic equation:
\[
(x - 3)^2 = 0
\]
Setting the factor to zero gives:
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Now that we have \( x = 3 \), we can substitute this value back into either of the original equations to find \( y \). We will use the second equation:
\[
y = -x - 6
\]
Substituting \( x = 3 \):
\[
y = -3 - 6 = -9
\]
Thus, we have the solution:
\[
(x, y) = (3, -9)
\]
Therefore, the solution to the system of equations is:
**A. The solution(s) is(are) (3, -9).**
1. \( y = x^2 - 7x + 3 \)
2. \( y = -x - 6 \)
We can set the two equations equal to each other since they both equal \( y \):
\[
x^2 - 7x + 3 = -x - 6
\]
Now, let's rearrange this equation to form a standard quadratic equation. First, move all terms to one side:
\[
x^2 - 7x + 3 + x + 6 = 0
\]
This simplifies to:
\[
x^2 - 6x + 9 = 0
\]
Next, we can factor this quadratic equation:
\[
(x - 3)^2 = 0
\]
Setting the factor to zero gives:
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Now that we have \( x = 3 \), we can substitute this value back into either of the original equations to find \( y \). We will use the second equation:
\[
y = -x - 6
\]
Substituting \( x = 3 \):
\[
y = -3 - 6 = -9
\]
Thus, we have the solution:
\[
(x, y) = (3, -9)
\]
Therefore, the solution to the system of equations is:
**A. The solution(s) is(are) (3, -9).**