Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point) Responses (−4,26) left parenthesis negative 4 comma 26 right parenthesis (−3,19) and (−4,26) left parenthesis negative 3 comma 19 right parenthesis and left parenthesis negative 4 comma 26 right parenthesis (−3,19) left parenthesis negative 3 comma 19 right parenthesis (3,−23) and (4,−30)
To solve the system using the elimination method, we can set the expressions for y equal to each other:
x^2 + 10 = -7x - 2
Now, rearrange the equation to standard quadratic form:
x^2 + 7x + 12 = 0
We can factor the quadratic:
(x + 3)(x + 4) = 0
Setting each factor equal to zero, we find two possible values for x:
x + 3 = 0 --> x = -3
x + 4 = 0 --> x = -4
Substituting these x-values back into either equation, we can find the corresponding y-values:
For x = -3:
y = (-3)^2 + 10 = 9 + 10 = 19
For x = -4:
y = (-4)^2 + 10 = 16 + 10 = 26
Therefore, the solution to the system is (-3, 19) and (-4, 26).