To solve the linear-quadratic system using the substitution method, we need to substitute one equation into the other and solve for the variable.
In this case, let's substitute the equation y = 3x - 4 into the equation y = x^2 - 4x + 8:
3x - 4 = x^2 - 4x + 8
Combining like terms:
0 = x^2 - 2x + 12
Now we have a quadratic equation. To solve it, we can factor it or use the quadratic formula. Let's factor it:
0 = (x - 4)(x + 3)
Setting each factor equal to zero:
x - 4 = 0 or x + 3 = 0
Solving for x:
x = 4 or x = -3
Now that we have the x-values, we can substitute them back into one of the original equations to find the corresponding y-values. Let's use the equation y = 3x - 4:
For x = 4: y = 3(4) - 4 = 12 - 4 = 8
For x = -3: y = 3(-3) - 4 = -9 - 4 = -13
Therefore, the solution to the linear-quadratic system is (4, 8) and (-3, -13).
The correct answer is: (−4,−16) and (−3,−13)
Which of the following uses the substitution method to solve linear-quadratic system y=x^2−4x+8 , y=3x−4 ?(1 point)
Responses
(4,8) and (3,5)
(−4,−16) and (−3,−13)
(3,5)
(4,8)
1 answer