((x-2)+(x)+(3x+1))=((2x-5)+(x+4)+(6X-7))

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Solution for ((x-2)+(x)+(3x+1))=((2x-5)+(x+4)+(6X-7)) equation: 



((x-2)+(x)+(3x+1))=((2x-5)+(x+4)+(6x-7))

We move all terms to the left:

((x-2)+(x)+(3x+1))-(((2x-5)+(x+4)+(6x-7)))=0



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

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

x+x+3x-2+1

We add all the numbers together, and all the variables

5x-1

Back to the equation:

+(5x-1)





We calculate terms in parentheses: -(((2x-5)+(x+4)+(6x-7))), so:

((2x-5)+(x+4)+(6x-7))



We calculate terms in parentheses: +((2x-5)+(x+4)+(6x-7)), so:

(2x-5)+(x+4)+(6x-7)

We get rid of parentheses

2x+x+6x-5+4-7

We add all the numbers together, and all the variables

9x-8

Back to the equation:

+(9x-8)



We get rid of parentheses

9x-8

Back to the equation:

-(9x-8)



We get rid of parentheses

5x-9x-1+8=0

We add all the numbers together, and all the variables

-4x+7=0

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

-4x=-7

x=-7/-4

x=1+3/4

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