(2x+3)+(2x+3)+(10x-5)=(3x-0.5)+(3x-0.5)+(5x-2)+(5x-2)

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Solution for (2x+3)+(2x+3)+(10x-5)=(3x-0.5)+(3x-0.5)+(5x-2)+(5x-2) equation: 



(2x+3)+(2x+3)+(10x-5)=(3x-0.5)+(3x-0.5)+(5x-2)+(5x-2)

We move all terms to the left:

(2x+3)+(2x+3)+(10x-5)-((3x-0.5)+(3x-0.5)+(5x-2)+(5x-2))=0

We get rid of parentheses

2x+2x+10x-((3x-0.5)+(3x-0.5)+(5x-2)+(5x-2))+3+3-5=0



We calculate terms in parentheses: -((3x-0.5)+(3x-0.5)+(5x-2)+(5x-2)), so:

(3x-0.5)+(3x-0.5)+(5x-2)+(5x-2)

We get rid of parentheses

3x+3x+5x+5x-0.5-0.5-2-2

We add all the numbers together, and all the variables

16x-5

Back to the equation:

-(16x-5)



We add all the numbers together, and all the variables

14x-(16x-5)+1=0

We get rid of parentheses

14x-16x+5+1=0

We add all the numbers together, and all the variables

-2x+6=0

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

-2x=-6

x=-6/-2

x=+3

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