(3x)/2+(2x/3)=(1+3x)/2

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Solution for (3x)/2+(2x/3)=(1+3x)/2 equation: 



(3x)/2+(2x/3)=(1+3x)/2

We move all terms to the left:

(3x)/2+(2x/3)-((1+3x)/2)=0

We add all the numbers together, and all the variables

3x/2+(+2x/3)-((3x+1)/2)=0

We get rid of parentheses

3x/2+2x/3-((3x+1)/2)=0

We calculate fractions


16x^2/()+9x/()+(-((3x+1)*3)/()=0



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

-((3x+1)*3)/(

We multiply all the terms by the denominator


-((3x+1)*3)



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

(3x+1)*3

We multiply parentheses

9x+3

Back to the equation:

-(9x+3)



We get rid of parentheses

-9x-3

Back to the equation:

+(-9x-3)



We get rid of parentheses

16x^2/()+9x/()-9x-3=0

We multiply all the terms by the denominator


16x^2+9x-9x*()-3*()=0

We add all the numbers together, and all the variables

16x^2+9x-9x*()=0

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