r+(2r-1)+2(3-r)=(4r+1)

Solution for r+(2r-1)+2(3-r)=(4r+1) equation:

r+(2r-1)+2(3-r)=(4r+1)

We move all terms to the left:

r+(2r-1)+2(3-r)-((4r+1))=0

We add all the numbers together, and all the variables

r+(2r-1)+2(-1r+3)-((4r+1))=0

We multiply parentheses

r+(2r-1)-2r-((4r+1))+6=0

We get rid of parentheses

r+2r-2r-((4r+1))-1+6=0


We calculate terms in parentheses: -((4r+1)), so:

(4r+1)

We get rid of parentheses

4r+1

Back to the equation:

-(4r+1)


We add all the numbers together, and all the variables

r-(4r+1)+5=0

We get rid of parentheses

r-4r-1+5=0

We add all the numbers together, and all the variables

-3r+4=0

We move all terms containing r to the left, all other terms to the right

-3r=-4

r=-4/-3

r=1+1/3