5(r+3)=3r+35

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Solution for 5(r+3)=3r+35 equation: 



5(r+3)=3r+35

We move all terms to the left:

5(r+3)-(3r+35)=0

We multiply parentheses

5r-(3r+35)+15=0

We get rid of parentheses

5r-3r-35+15=0

We add all the numbers together, and all the variables

2r-20=0

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

2r=20

r=20/2

r=10

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