Solve the system of equations algebraically.

{Y=(x-2)^2+2
{Y+4=3x

(5, 11) and (2, 2)

Step-by-step explanation:

y = (x-2)² + 2

y + 4 = 3x

(x-2)² + 2 + 4 = 3x

x² - 4x + 4 + 6 = 3x

x² - 7x + 10 = 0

(x - 5)(x - 2) = 0

x - 5 = 0, x = 5

x - 2 = 0, x = 2

y = (5-2)² + 2 = 11

(5, 11)

y = (2-2)² + 2 = 2

(2, 2)

Step-by-step explanation:

Hello, please consider the following.

We want to solve this system of equations.

This is equivalent to (subtract 4 from the second equation).

Then, we can write y = y, meaning:

For x = 2, y =0+2=2 (from the first equation) and for x = 5 y=3*5-4=15-4=11 (from the second equation)

So the solutions are (2,2) and (5,11)

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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