To determine the solution set of the system of equations [tex]\( y = x^2 - 6x + 12 \)[/tex] and [tex]\( y = 2x - 4 \)[/tex] algebraically, follow these steps:
### Step 1: Set the equations equal to each other
Given the equations [tex]\( y = x^2 - 6x + 12 \)[/tex] and [tex]\( y = 2x - 4 \)[/tex], set them equal to each other to find the values of [tex]\( x \)[/tex] where the two equations intersect:
[tex]\[ x^2 - 6x + 12 = 2x - 4 \][/tex]
### Step 2: Rearrange into a standard form quadratic equation
To create a standard form quadratic equation (i.e., [tex]\( ax^2 + bx + c = 0 \)[/tex]), move all terms to one side of the equation:
[tex]\[ x^2 - 6x + 12 - 2x + 4 = 0 \][/tex]
Combine like terms:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]
### Step 3: Solve the quadratic equation
Solve the quadratic equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex].
### Roots of the quadratic equation
The roots of the equation [tex]\( x^2 - 8x + 16 = 0 \)[/tex] will give us the x-values where the solutions occur. In this case, the roots are:
[tex]\[ x = 4 \][/tex]
(since)[tex]$(x-4)^2=0$[/tex]
### Step 4: Find the corresponding y-values
Substitute [tex]\( x = 4 \)[/tex] back into either of the original equations to find the corresponding y-values.
Using the equation [tex]\( y = 2x - 4 \)[/tex]:
[tex]\[ y = 2(4) - 4 = 8 - 4 = 4 \][/tex]
This means the solution set for the system of equations [tex]\( y = x^2 - 6x + 12 \)[/tex] and [tex]\( y = 2x - 4 \)[/tex] is:
[tex]\[ (4, 4) \][/tex]
Therefore, the correct solution(s) of this system of equations are:
[tex]\[ (4, 4) \][/tex]
