2(r+5)+20=3(r-4)+5r

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Solution for 2(r+5)+20=3(r-4)+5r equation: 



2(r+5)+20=3(r-4)+5r

We move all terms to the left:

2(r+5)+20-(3(r-4)+5r)=0

We multiply parentheses

2r-(3(r-4)+5r)+10+20=0



We calculate terms in parentheses: -(3(r-4)+5r), so:

3(r-4)+5r

We add all the numbers together, and all the variables

5r+3(r-4)

We multiply parentheses

5r+3r-12

We add all the numbers together, and all the variables

8r-12

Back to the equation:

-(8r-12)



We add all the numbers together, and all the variables

2r-(8r-12)+30=0

We get rid of parentheses

2r-8r+12+30=0

We add all the numbers together, and all the variables

-6r+42=0

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

-6r=-42

r=-42/-6

r=+7

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