360=3x+(2x/10)+(2x+5)+(2x-15)

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Solution for 360=3x+(2x/10)+(2x+5)+(2x-15) equation: 



360=3x+(2x/10)+(2x+5)+(2x-15)

We move all terms to the left:

360-(3x+(2x/10)+(2x+5)+(2x-15))=0



Domain of the equation: 10)+(2x+5)+(2x-15))!=0

x∈R



We add all the numbers together, and all the variables

-(3x+(+2x/10)+(2x+5)+(2x-15))+360=0

We multiply all the terms by the denominator


-(3x+(+2x+360*10)+(2x+5)+(2x-15))=0



We calculate terms in parentheses: -(3x+(+2x+360*10)+(2x+5)+(2x-15)), so:

3x+(+2x+360*10)+(2x+5)+(2x-15)

We add all the numbers together, and all the variables

3x+(2x+3600)+(2x+5)+(2x-15)

We get rid of parentheses

3x+2x+2x+2x+3600+5-15

We add all the numbers together, and all the variables

9x+3590

Back to the equation:

-(9x+3590)



We get rid of parentheses

-9x-3590=0

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

-9x=3590

x=3590/-9

x=-398+8/9

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