2(x-3)+1/7(7+2x)=1/2+3(x-1)

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Solution for 2(x-3)+1/7(7+2x)=1/2+3(x-1) equation: 



2(x-3)+1/7(7+2x)=1/2+3(x-1)

We move all terms to the left:

2(x-3)+1/7(7+2x)-(1/2+3(x-1))=0



Domain of the equation: 7(7+2x)!=0

x∈R





Domain of the equation: 2+3(x-1))!=0

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

3(x-1))!=-2

x∈R



We add all the numbers together, and all the variables

2(x-3)+1/7(2x+7)-(1/2+3(x-1))=0

We multiply parentheses

2x+1/7(2x+7)-(1/2+3(x-1))-6=0

We calculate fractions


2x+(3(x-1))+1*2)/31x+(-7x2/31x-6=0



We calculate terms in parentheses: +(3(x-1)), so:

3(x-1)

We multiply parentheses

3x-3

Back to the equation:

+(3x-3)



We get rid of parentheses

2x+3x+1*2)/31x+(-7x2/31x-3-6=0

We calculate fractions

We do not support expression: x^3

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