Verified answer

y = x2 − 4x + 3y

-2y = x^2 -4x

y = -1/2x^2 + 2x

y = x - 1

-1/2x^2 + 2x = x - 1

x^2 + 4x = 2x - 2

x^2 + 2x + 2 = 0

solve for x using quadratic formula with:

a=1

b=2

c=2

{-2 +/- [(2)^2 - 4(1)(2)]^(1/2)} / (2)(1)

discriminant is negative so the roots are imaginary numbers

Set the two equations equivalent to a minimum of one yet another x2-4x+4=-x2+6x-8 flow each and all the variables to the left area by skill of including to the two aspects 2x2-10x+12=0 Divide the two aspects by skill of two x2-5x+6=0 element the quadratic equation. (x-3)(x-2)=0 resolve x-3=0 and x-2=0 to your 2 solutions. x=3, x=2

y = x^2 − 4x + 3

if y = x-1

then

x^2 − 4x + 3 = y

x^2 − 4x + 3 = x - 1

x^2 - 5x + 4 = 0

(x-4)(x-1) = 0

x = 4 and x = 1

y = x^2 − 4x + 3

y = x − 1

x^2 − 4x + 3 = x - 1

x^2 - 5x + 4 = 0

(x-4)(x-1) = 0

x = 4 and x = 1