3(x-1)-8=41(1+x)+5

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Solution for 3(x-1)-8=41(1+x)+5 equation:



3(x-1)-8=41(1+x)+5
We move all terms to the left:
3(x-1)-8-(41(1+x)+5)=0
We add all the numbers together, and all the variables
3(x-1)-(41(x+1)+5)-8=0
We multiply parentheses
3x-(41(x+1)+5)-3-8=0
We calculate terms in parentheses: -(41(x+1)+5), so:
41(x+1)+5
We multiply parentheses
41x+41+5
We add all the numbers together, and all the variables
41x+46
Back to the equation:
-(41x+46)
We add all the numbers together, and all the variables
3x-(41x+46)-11=0
We get rid of parentheses
3x-41x-46-11=0
We add all the numbers together, and all the variables
-38x-57=0
We move all terms containing x to the left, all other terms to the right
-38x=57
x=57/-38
x=-1+1/2

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