(3/5)(y-2)=1-3y

Solution for (3/5)(y-2)=1-3y equation:

(3/5)(y-2)=1-3y

We move all terms to the left:

(3/5)(y-2)-(1-3y)=0


Domain of the equation: 5)(y-2)!=0

y∈R


We add all the numbers together, and all the variables

(+3/5)(y-2)-(-3y+1)=0

We get rid of parentheses

(+3/5)(y-2)+3y-1=0

We multiply parentheses ..

(+3y^2+3/5*-2)+3y-1=0

We multiply all the terms by the denominator


(+3y^2+3+3y*5*-2)-1*5*-2)=0

We add all the numbers together, and all the variables

(+3y^2+3+3y*5*-2)=0

We get rid of parentheses

3y^2+3y*5*+3-2=0

We add all the numbers together, and all the variables

3y^2+3y*5*+1=0

Wy multiply elements

3y^2+15y^2+1=0

We add all the numbers together, and all the variables

18y^2+1=0

a = 18; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·18·1
Δ = -72
Delta is less than zero, so there is no solution for the equation