(2x-3)/8+x=(2x+3)/2=6

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Solution for (2x-3)/8+x=(2x+3)/2=6 equation: 



(2x-3)/8+x=(2x+3)/2=6

We move all terms to the left:

(2x-3)/8+x-((2x+3)/2)=0

We add all the numbers together, and all the variables

x+(2x-3)/8-((2x+3)/2)=0

We calculate fractions


x+2x/()+(-((2x+3)*8)/()=0



We calculate terms in parentheses: +(-((2x+3)*8)/(), so:

-((2x+3)*8)/(

We multiply all the terms by the denominator


-((2x+3)*8)



We calculate terms in parentheses: -((2x+3)*8), so:

(2x+3)*8

We multiply parentheses

16x+24

Back to the equation:

-(16x+24)



We get rid of parentheses

-16x-24

Back to the equation:

+(-16x-24)



We get rid of parentheses

x+2x/()-16x-24=0

We multiply all the terms by the denominator


x*()+2x-16x*()-24*()=0

We add all the numbers together, and all the variables

2x+x*()-16x*()=0

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