9r+3(r-4)=10(r-5)+6

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Solution for 9r+3(r-4)=10(r-5)+6 equation: 



9r+3(r-4)=10(r-5)+6

We move all terms to the left:

9r+3(r-4)-(10(r-5)+6)=0

We multiply parentheses

9r+3r-(10(r-5)+6)-12=0



We calculate terms in parentheses: -(10(r-5)+6), so:

10(r-5)+6

We multiply parentheses

10r-50+6

We add all the numbers together, and all the variables

10r-44

Back to the equation:

-(10r-44)



We add all the numbers together, and all the variables

12r-(10r-44)-12=0

We get rid of parentheses

12r-10r+44-12=0

We add all the numbers together, and all the variables

2r+32=0

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

2r=-32

r=-32/2

r=-16

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