What are the solutions to the systems?
y=x^2+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
oh dear you people for 5 years ago really hate giving answers
Can anyone help me?
So D?
check for a typo. There are two solutions, but they are not listed.
To find the solutions to the system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
Let's start by setting the two equations equal to each other:
x^2 + 6x + 7 = -x + 13
Now, we can simplify the equation by moving all terms to one side:
x^2 + 7x - (-x) + 7 - 13 = 0
x^2 + 8x - x - 6 = 0
x^2 + 7x - 6 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, the equation can be factored as:
(x + 6)(x - 1) = 0
This gives us two possible values for x: x = -6 and x = 1.
Now, we substitute these values of x back into either of the original equations to find the corresponding values of y.
For x = -6:
y = (-6)^2 + 6(-6) + 7
= 36 - 36 + 7
= 7
So, one solution is (-6, 7).
For x = 1:
y = (1)^2 + 6(1) + 7
= 1 + 6 + 7
= 14
So, the other solution is (1, 14).
Therefore, the solutions to the system of equations are:
(-6, 7) and (1, 14).
Therefore, the correct answer is option A: (-1,14) and (6,7).
substituting
-x + 13 = x² + 6 x + 7
0 = x² + 7 x - 6
none of the answers are correct