What are the solutions to the system?
y = x2 – 6x + 7
y = –x + 13
A. (–1, 14) and (6, 7)
B. (–1, 7) and (6, 14)
C. (–1, 13) and (6, 7)
D. no solution
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Apologies for the confusion. The system of equations given is:
y = x^2 - 6x + 7
y = -x + 13
To find the solutions, we can set the two equations equal to each other:
x^2 - 6x + 7 = -x + 13
Simplifying this equation gives:
x^2 - 5x - 6 = 0
Factoring this equation gives:
(x - 6)(x + 1) = 0
Therefore, the solutions are x = -1 and x = 6.
To find the corresponding y-values, we can substitute these values of x back into either of the original equations. Using the second equation:
y = -x + 13
When x = -1, y = -(-1) + 13 = 14.
When x = 6, y = -(6) + 13 = 7.
Therefore, the solutions to the system are (-1, 14) and (6, 7), which is answer choice A.
To find the solutions to the system of equations:
1. Set the two equations equal to each other:
x^2 - 6x + 7 = -x + 13
2. Rearrange the equation to make it a quadratic equation equal to zero:
x^2 - 6x + x + 7 - 13 = 0
x^2 - 5x - 6 = 0
3. Factor the quadratic equation:
(x - 6)(x + 1) = 0
4. Set each factor equal to zero and solve for x:
x - 6 = 0 or x + 1 = 0
x = 6 or x = -1
5. Substitute the values of x into either of the original equations to find the corresponding values of y.
For x = 6:
y = (6)^2 - 6(6) + 7
y = 36 - 36 + 7
y = 7
For x = -1:
y = (-1)^2 - 6(-1) + 7
y = 1 + 6 + 7
y = 14
Therefore, the solution to the system of equations is (x, y) = (-1, 14) and (6, 7).
The correct answer is option A: (–1, 14) and (6, 7).
To find the solutions to the system, we need to solve the given equations simultaneously by finding the values of x and y that satisfy both equations.
Let's start by equating the two equations:
x^2 - 6x + 7 = -x + 13
Now, let's simplify the equation:
x^2 - 6x + x + 7 - 13 = 0
x^2 - 5x - 6 = 0
Now, we have a quadratic equation. To find the solutions, we can either factorize the equation or use the quadratic formula.
Factoring the equation:
(x - 6)(x + 1) = 0
Now, we can use the zero product property, which states that if a product of factors is equal to zero, then at least one of the factors must be equal to zero.
So, either (x - 6) = 0 or (x + 1) = 0.
Solving for x, we get:
x - 6 = 0 or x + 1 = 0
x = 6 or x = -1
Now that we have the values of x, we can substitute them back into one of the original equations to find the corresponding values of y.
Let's substitute x = 6 into the first equation:
y = (6)^2 - 6(6) + 7
y = 36 - 36 + 7
y = 7 + 7
y = 14
So one solution is (x, y) = (6, 14).
Now, let's substitute x = -1 into the first equation:
y = (-1)^2 - 6(-1) + 7
y = 1 + 6 + 7
y = 14
So another solution is (x, y) = (-1, 14).
Therefore, the solutions to the system of equations are (x, y) = (6, 14) and (-1, 14).
The correct answer is A. (–1, 14) and (6, 7).