Question one use the elimination method to solve the linear quadratic system why = x2 + 10, y = -7x - 2 is it -4, 26 -3, 19 -3, 19 and -4, 26 or is it 3, -23 and 4, -30
Question 2 use the elimination method to solve the linear quadratic system y = x2 - 2x - 2, y = 2x - 6 is it two common negative 2 -2, -10 2, -2 and -2, 2 or -2, -2 and -2, 2
To solve the linear-quadratic system y = x^2 + 10 and y = -7x - 2 using the elimination method, you can substitute the value of y from the second equation into the first equation:
-7x - 2 = x^2 + 10
Rearranging this equation, we get:
x^2 + 7x + 12 = 0
Factoring this equation, we get:
(x + 3)(x + 4) = 0
Setting each factor equal to zero, we get:
x + 3 = 0 or x + 4 = 0
Solving these equations, we find x = -3 or x = -4.
Then, substituting these values back into either of the original equations, we can determine the corresponding values of y:
When x = -3, y = (-7)(-3) - 2 = 19.
When x = -4, y = (-7)(-4) - 2 = 26.
Therefore, the solution to the linear-quadratic system is (-3, 19) and (-4, 26).
Answer to question one: The correct answer is (-3, 19) and (-4, 26).
To solve the linear-quadratic system y = x^2 - 2x - 2 and y = 2x - 6 using the elimination method, you can substitute the value of y from the second equation into the first equation:
2x - 6 = x^2 - 2x - 2
Rearranging this equation, we get:
x^2 - 4x + 4 = 0
Factoring this equation, we get:
(x - 2)(x - 2) = 0
Setting the factor equal to zero, we get:
x - 2 = 0
Solving this equation, we find x = 2.
Then, substituting this value back into either of the original equations, we can determine the corresponding value of y:
When x = 2, y = (2)(2) - 6 = -2.
Therefore, the solution to the linear-quadratic system is (2, -2).
Answer to question two: The correct answer is (2, -2).