2(r+5)-3=3(r-8)+20

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Solution for 2(r+5)-3=3(r-8)+20 equation:



2(r+5)-3=3(r-8)+20
We move all terms to the left:
2(r+5)-3-(3(r-8)+20)=0
We multiply parentheses
2r-(3(r-8)+20)+10-3=0
We calculate terms in parentheses: -(3(r-8)+20), so:
3(r-8)+20
We multiply parentheses
3r-24+20
We add all the numbers together, and all the variables
3r-4
Back to the equation:
-(3r-4)
We add all the numbers together, and all the variables
2r-(3r-4)+7=0
We get rid of parentheses
2r-3r+4+7=0
We add all the numbers together, and all the variables
-1r+11=0
We move all terms containing r to the left, all other terms to the right
-r=-11
r=-11/-1
r=+11

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