((b+4)+b+(b-1))/3=97

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Solution for ((b+4)+b+(b-1))/3=97 equation: 



((b+4)+b+(b-1))/3=97

We move all terms to the left:

((b+4)+b+(b-1))/3-(97)=0

We multiply all the terms by the denominator


((b+4)+b+(b-1))-97*3=0



We calculate terms in parentheses: +((b+4)+b+(b-1)), so:

(b+4)+b+(b-1)

We add all the numbers together, and all the variables

b+(b+4)+(b-1)

We get rid of parentheses

b+b+b+4-1

We add all the numbers together, and all the variables

3b+3

Back to the equation:

+(3b+3)



We add all the numbers together, and all the variables

(3b+3)-291=0

We get rid of parentheses

3b+3-291=0

We add all the numbers together, and all the variables

3b-288=0

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

3b=288

b=288/3

b=96

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