To use the substitution method, we will substitute the value of y from the second equation into the first equation.
We have the second equation: y = 3x - 4
Substituting this into the first equation, we get: 3x - 4 = x^2 - 4x + 8
Rearranging this equation, we get: x^2 - 7x + 12 = 0
Now we can factor this equation: (x - 3)(x - 4) = 0
Setting each factor equal to zero, we get: x - 3 = 0 and x - 4 = 0
This gives us two possible values for x: x = 3 and x = 4
Substituting these values back into the second equation, we can find the corresponding y values:
For x = 3, we have: y = 3(3) - 4 = 9 - 4 = 5
For x = 4, we have: y = 3(4) - 4 = 12 - 4 = 8
Therefore, the solution to the linear-quadratic system using the substitution method is the points (3, 5) and (4, 8). Option (3,5) and option (4,8) both represent this solution.
Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8 , y=3x−4 ?(1 point)Responses(4,8)left parenthesis 4 comma 8 right parenthesis(4,8) and (3,5)left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis(3,5)left parenthesis 3 comma 5 right parenthesis(−4,−16) and (−3,−13)
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