(4x+3)+(4x+2)+(4x)=(80x+3)

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Solution for (4x+3)+(4x+2)+(4x)=(80x+3) equation: 



(4x+3)+(4x+2)+(4x)=(80x+3)

We move all terms to the left:

(4x+3)+(4x+2)+(4x)-((80x+3))=0

We add all the numbers together, and all the variables

4x+(4x+3)+(4x+2)-((80x+3))=0

We get rid of parentheses

4x+4x+4x-((80x+3))+3+2=0



We calculate terms in parentheses: -((80x+3)), so:

(80x+3)

We get rid of parentheses

80x+3

Back to the equation:

-(80x+3)



We add all the numbers together, and all the variables

12x-(80x+3)+5=0

We get rid of parentheses

12x-80x-3+5=0

We add all the numbers together, and all the variables

-68x+2=0

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

-68x=-2

x=-2/-68

x=1/34

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