(4x-3)/8+x=(4x+3)/4+2

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Solution for (4x-3)/8+x=(4x+3)/4+2 equation: 



(4x-3)/8+x=(4x+3)/4+2

We move all terms to the left:

(4x-3)/8+x-((4x+3)/4+2)=0

We add all the numbers together, and all the variables

x+(4x-3)/8-((4x+3)/4+2)=0

We calculate fractions


x+(16x-12)/()+(-((4x+3)*8)/()=0



We calculate terms in parentheses: +(-((4x+3)*8)/(), so:

-((4x+3)*8)/(

We multiply all the terms by the denominator


-((4x+3)*8)



We calculate terms in parentheses: -((4x+3)*8), so:

(4x+3)*8

We multiply parentheses

32x+24

Back to the equation:

-(32x+24)



We get rid of parentheses

-32x-24

Back to the equation:

+(-32x-24)



We get rid of parentheses

x+(16x-12)/()-32x-24=0

We multiply all the terms by the denominator


x*()+(16x-12)-32x*()-24*()=0

We add all the numbers together, and all the variables

x*()+(16x-12)-32x*()=0

We get rid of parentheses

x*()+16x-32x*()-12=0

We add all the numbers together, and all the variables

16x+x*()-32x*()-12=0

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

16x+x*()-32x*()=12

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