Use the elimination method to solve the linear-quadratic system y=x2+10 , y=−7x−2 .(1 point) Responses (3,−23) and (4,−30) left parenthesis 3 comma negative 23 right parenthesis and left parenthesis 4 comma negative 30 right parenthesis (−3,19) left parenthesis negative 3 comma 19 right parenthesis (−4,26) left parenthesis negative 4 comma 26 right parenthesis (−3,19) and (−4,26)
To solve the given linear-quadratic system using the elimination method, we can substitute one equation into the other and solve for the variable.
From the second equation, we have y = -7x - 2. Substituting this value of y into the first equation, we get:
x^2 + 10 = -7x - 2
Rearranging this equation, we have:
x^2 + 7x + 12 = 0
Factoring the quadratic equation, we have:
(x + 3)(x + 4) = 0
Setting each factor to zero, we get:
x + 3 = 0 or x + 4 = 0
Solving for x in both cases, we find:
x = -3 or x = -4
Substituting these values back into the second equation, we can find the corresponding y-values.
For x = -3:
y = -7x - 2
y = -7(-3) - 2
y = 21 - 2
y = 19
Therefore, one solution is (-3, 19).
For x = -4:
y = -7x - 2
y = -7(-4) - 2
y = 28 - 2
y = 26
Therefore, another solution is (-4, 26).
Hence, the correct response is (-3, 19) and (-4, 26).