((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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