(2x-5)/8+x=(2x+5)/2+1

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Solution for (2x-5)/8+x=(2x+5)/2+1 equation: 



(2x-5)/8+x=(2x+5)/2+1

We move all terms to the left:

(2x-5)/8+x-((2x+5)/2+1)=0

We add all the numbers together, and all the variables

x+(2x-5)/8-((2x+5)/2+1)=0

We calculate fractions


x+(4x-10)/()+(-((2x+5)*8)/()=0



We calculate terms in parentheses: +(-((2x+5)*8)/(), so:

-((2x+5)*8)/(

We multiply all the terms by the denominator


-((2x+5)*8)



We calculate terms in parentheses: -((2x+5)*8), so:

(2x+5)*8

We multiply parentheses

16x+40

Back to the equation:

-(16x+40)



We get rid of parentheses

-16x-40

Back to the equation:

+(-16x-40)



We get rid of parentheses

x+(4x-10)/()-16x-40=0

We multiply all the terms by the denominator


x*()+(4x-10)-16x*()-40*()=0

We add all the numbers together, and all the variables

x*()+(4x-10)-16x*()=0

We get rid of parentheses

x*()+4x-16x*()-10=0

We add all the numbers together, and all the variables

4x+x*()-16x*()-10=0

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

4x+x*()-16x*()=10

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