540=120+(2x)+(5x-12)+(10x-8)+(3x)

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Solution for 540=120+(2x)+(5x-12)+(10x-8)+(3x) equation: 



540=120+(2x)+(5x-12)+(10x-8)+(3x)

We move all terms to the left:

540-(120+(2x)+(5x-12)+(10x-8)+(3x))=0



We calculate terms in parentheses: -(120+2x+(5x-12)+(10x-8)+3x), so:

120+2x+(5x-12)+(10x-8)+3x

determiningTheFunctionDomain
2x+(5x-12)+(10x-8)+3x+120

We add all the numbers together, and all the variables

5x+(5x-12)+(10x-8)+120

We get rid of parentheses

5x+5x+10x-12-8+120

We add all the numbers together, and all the variables

20x+100

Back to the equation:

-(20x+100)



We get rid of parentheses

-20x-100+540=0

We add all the numbers together, and all the variables

-20x+440=0

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

-20x=-440

x=-440/-20

x=+22

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