1/5(x-3)=5x-4(x-1)+41/5

Solution for 1/5(x-3)=5x-4(x-1)+41/5 equation:

1/5(x-3)=5x-4(x-1)+41/5

We move all terms to the left:

1/5(x-3)-(5x-4(x-1)+41/5)=0


Domain of the equation: 5(x-3)!=0

x∈R


We calculate fractions


()/5x+(-(x+205xx)/5x=0

We add all the numbers together, and all the variables

()/5x+(-(+x+205xx)/5x=0

We multiply all the terms by the denominator


(-(+x+205xx)+()=0


We calculate terms in parentheses: +(-(+x+205xx)+(), so:

-(+x+205xx)+(

We add all the numbers together, and all the variables

-(+x+205xx)

We get rid of parentheses

-x-205xx

We add all the numbers together, and all the variables

-1x-205xx

Back to the equation:

+(-1x-205xx)


We get rid of parentheses

-1x-205xx=0