3501=3(x+62)+4X+2(2/5X)

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Solution for 3501=3(x+62)+4X+2(2/5X) equation: 



3501=3(x+62)+4x+2(2/5x)

We move all terms to the left:

3501-(3(x+62)+4x+2(2/5x))=0



Domain of the equation: 5x))!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

-(3(x+62)+4x+2(+2/5x))+3501=0

We multiply all the terms by the denominator


-(3(x+62)+4x+2(+2+3501*5x))=0



We calculate terms in parentheses: -(3(x+62)+4x+2(+2+3501*5x)), so:

3(x+62)+4x+2(+2+3501*5x)

We add all the numbers together, and all the variables

3(x+62)+4x+2(3501*5x+2)

We add all the numbers together, and all the variables

4x+3(x+62)+2(3501*5x+2)

We multiply parentheses

4x+3x+35010x+186+4

We add all the numbers together, and all the variables

35017x+190

Back to the equation:

-(35017x+190)



We get rid of parentheses

-35017x-190=0

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

-35017x=190

x=190/-35017

x=-10/1843

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