(8y/9)=(1/2)-(y-12)

Solution for (8y/9)=(1/2)-(y-12) equation:

(8y/9)=(1/2)-(y-12)

We move all terms to the left:

(8y/9)-((1/2)-(y-12))=0


Domain of the equation: 2)-(y-12))!=0

y∈R


We add all the numbers together, and all the variables

(+8y/9)-((+1/2)-(y-12))=0

We get rid of parentheses

8y/9-((+1/2)-(y-12))=0

We calculate fractions


16y^2/18y+()/18y=0

We multiply all the terms by the denominator


16y^2+()=0

We add all the numbers together, and all the variables

16y^2=0

a = 16; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·16·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{32}=0$