Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point)
Responses
(−3,19) and (−4,26)
left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis
(−4,26)
left parenthesis negative 4 comma 26 right parenthesis
(3,−23) and (4,−30)
left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis
(−3,19)
To solve this system using the elimination method, we can set the two equations equal to each other:
x^2 + 10 = -7x - 2
Rearranging this equation to set it equal to zero:
x^2 + 7x + 12 = 0
Now we can factor this equation:
(x + 3)(x + 4) = 0
This gives us two possible solutions for x: x = -3 and x = -4.
Using these values of x to find the corresponding values of y:
For x = -3:
y = (-3)^2 + 10 = 9 + 10 = 19
For x = -4:
y = (-4)^2 + 10 = 16 + 10 = 26
Therefore, the solutions to the linear-quadratic system are (-3, 19) and (-4, 26).