540=139+(3x)+(5x-10)+(10x-9)+(3x)

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Solution for 540=139+(3x)+(5x-10)+(10x-9)+(3x) equation: 



540=139+(3x)+(5x-10)+(10x-9)+(3x)

We move all terms to the left:

540-(139+(3x)+(5x-10)+(10x-9)+(3x))=0



We calculate terms in parentheses: -(139+3x+(5x-10)+(10x-9)+3x), so:

139+3x+(5x-10)+(10x-9)+3x

determiningTheFunctionDomain
3x+(5x-10)+(10x-9)+3x+139

We add all the numbers together, and all the variables

6x+(5x-10)+(10x-9)+139

We get rid of parentheses

6x+5x+10x-10-9+139

We add all the numbers together, and all the variables

21x+120

Back to the equation:

-(21x+120)



We get rid of parentheses

-21x-120+540=0

We add all the numbers together, and all the variables

-21x+420=0

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

-21x=-420

x=-420/-21

x=+20

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