(3x-5)+(19-x)+(2x)=(3x-5)+(19-x)+(2x)

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Solution for (3x-5)+(19-x)+(2x)=(3x-5)+(19-x)+(2x) equation: 



(3x-5)+(19-x)+(2x)=(3x-5)+(19-x)+(2x)

We move all terms to the left:

(3x-5)+(19-x)+(2x)-((3x-5)+(19-x)+(2x))=0

We add all the numbers together, and all the variables

(3x-5)+(-1x+19)+2x-((3x-5)+(-1x+19)+2x)=0

We add all the numbers together, and all the variables

2x+(3x-5)+(-1x+19)-((3x-5)+(-1x+19)+2x)=0

We get rid of parentheses

2x+3x-1x-((3x-5)+(-1x+19)+2x)-5+19=0



We calculate terms in parentheses: -((3x-5)+(-1x+19)+2x), so:

(3x-5)+(-1x+19)+2x

We add all the numbers together, and all the variables

2x+(3x-5)+(-1x+19)

We get rid of parentheses

2x+3x-1x-5+19

We add all the numbers together, and all the variables

4x+14

Back to the equation:

-(4x+14)



We add all the numbers together, and all the variables

4x-(4x+14)+14=0

We get rid of parentheses

4x-4x-14+14=0

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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