3(r+1)=(2(3-2r)-3(3-r))=2(r+5)-4

Solution for 3(r+1)=(2(3-2r)-3(3-r))=2(r+5)-4 equation:

3(r+1)=(2(3-2r)-3(3-r))=2(r+5)-4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -((2(-2r+3)-3(-1r+3))), so:

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


We calculate terms in parentheses: +(2(-2r+3)-3(-1r+3)), so:

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

We multiply parentheses

-4r+3r+6-9

We add all the numbers together, and all the variables

-1r-3

Back to the equation:

+(-1r-3)


We get rid of parentheses

-1r-3

Back to the equation:

-(-1r-3)


We get rid of parentheses

3r+1r+3+3=0

We add all the numbers together, and all the variables

4r+6=0

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

4r=-6

r=-6/4

r=-1+1/2