73-(b+8)-2(3b+4)=6b-2(b-5)+69

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Solution for 73-(b+8)-2(3b+4)=6b-2(b-5)+69 equation:



73-(b+8)-2(3b+4)=6b-2(b-5)+69
We move all terms to the left:
73-(b+8)-2(3b+4)-(6b-2(b-5)+69)=0
We multiply parentheses
-(b+8)-6b-(6b-2(b-5)+69)-8+73=0
We get rid of parentheses
-b-6b-(6b-2(b-5)+69)-8-8+73=0
We calculate terms in parentheses: -(6b-2(b-5)+69), so:
6b-2(b-5)+69
We multiply parentheses
6b-2b+10+69
We add all the numbers together, and all the variables
4b+79
Back to the equation:
-(4b+79)
We add all the numbers together, and all the variables
-7b-(4b+79)+57=0
We get rid of parentheses
-7b-4b-79+57=0
We add all the numbers together, and all the variables
-11b-22=0
We move all terms containing b to the left, all other terms to the right
-11b=22
b=22/-11
b=-2

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