2(3y-4)=3(y-2/3)=

Solution for 2(3y-4)=3(y-2/3)= equation:

2(3y-4)=3(y-2/3)=

We move all terms to the left:

2(3y-4)-(3(y-2/3))=0

We add all the numbers together, and all the variables

2(3y-4)-(3(+y-2/3))=0

We multiply parentheses

6y-(3(+y-2/3))-8=0

We multiply all the terms by the denominator


6y*3))-(3(+y-2-8*3))=0

We add all the numbers together, and all the variables

6y*3))-(3(y-26))=0

We add all the numbers together, and all the variables

6y*3))-(3(y=0

Wy multiply elements

18y^2=0

a = 18; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·18·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{36}=0$