Solve the system y = (x - 2)^2 + 2, y + 4 = 3x.
Please show steps.
y = 3x-4 and y = x^2 -4x+4+2 equate them
x^2 - 4x +6 = 3x-4
x^2-7x+10 = 0
(x-5)(x-2) = 0 so x = 5 and 2
use these values of 'x' in one of the equations to compute the coresponding 'y' s
To start, we can set y from the second equation to 3x-4. Plugging this in to the first equation gives us 3x-4 = (x-2)^2+2.
From here, we can add 4 to both sides and get 3x = (x-2)^2+6
Multiplying out (x-2)^2, we get x^2-4x+4.
Now, our equation is 3x = x^2-4x+10. Subtracting 3x from both sides will give us 0 = x^2-7x+10. Can you solve the quadratic from here?