3501=3(x+62)+(4x)+(2*0,4x)

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Solution for 3501=3(x+62)+(4x)+(2*0,4x) equation: 



3501=3(x+62)+(4x)+(2*0.4x)

We move all terms to the left:

3501-(3(x+62)+(4x)+(2*0.4x))=0

We add all the numbers together, and all the variables

-(3(x+62)+4x+(+2*0.4x))+3501=0



We calculate terms in parentheses: -(3(x+62)+4x+(+2*0.4x)), so:

3(x+62)+4x+(+2*0.4x)

We add all the numbers together, and all the variables

4x+3(x+62)+(+2*0.4x)

We multiply parentheses

4x+3x+(+2*0.4x)+186

We get rid of parentheses

4x+3x+2*0.4x+186

We add all the numbers together, and all the variables

7x+2*0.4x+186

Wy multiply elements

7x+0x+186

We add all the numbers together, and all the variables

8x+186

Back to the equation:

-(8x+186)



We get rid of parentheses

-8x-186+3501=0

We add all the numbers together, and all the variables

-8x+3315=0

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

-8x=-3315

x=-3315/-8

x=414+3/8

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