38=(2x+1)+(2x+1)+(3x-2)+(3x-2)

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Solution for 38=(2x+1)+(2x+1)+(3x-2)+(3x-2) equation: 



38=(2x+1)+(2x+1)+(3x-2)+(3x-2)

We move all terms to the left:

38-((2x+1)+(2x+1)+(3x-2)+(3x-2))=0



We calculate terms in parentheses: -((2x+1)+(2x+1)+(3x-2)+(3x-2)), so:

(2x+1)+(2x+1)+(3x-2)+(3x-2)

We get rid of parentheses

2x+2x+3x+3x+1+1-2-2

We add all the numbers together, and all the variables

10x-2

Back to the equation:

-(10x-2)



We get rid of parentheses

-10x+2+38=0

We add all the numbers together, and all the variables

-10x+40=0

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

-10x=-40

x=-40/-10

x=+4

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