[2*(x+1)/3+(5-x)/2=(1-x)/2+2

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Solution for [2*(x+1)/3+(5-x)/2=(1-x)/2+2 equation: 



2(x+1)/3+(5-x)/2=(1-x)/2+2

We move all terms to the left:

2(x+1)/3+(5-x)/2-((1-x)/2+2)=0

We add all the numbers together, and all the variables

2(x+1)/3+(-1x+5)/2-((-1x+1)/2+2)=0

We calculate fractions


(2(x+1)*2*2)/()+(-3x+15)/()+(-((-1x+1)*3)/()=0



We calculate terms in parentheses: +(2(x+1)*2*2)/(), so:

2(x+1)*2*2)/(

We multiply all the terms by the denominator


2(x+1)*2*2)

We multiply parentheses

2x+

We add all the numbers together, and all the variables

2x

Back to the equation:

+(2x)





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

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

We multiply all the terms by the denominator


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



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

(-1x+1)*3

We multiply parentheses

-3x+3

Back to the equation:

-(-3x+3)



We get rid of parentheses

3x-3

Back to the equation:

+(3x-3)



We get rid of parentheses

2x+(-3x+15)/()+3x-3=0

We multiply all the terms by the denominator


2x*()+(-3x+15)+3x*()-3*()=0

We add all the numbers together, and all the variables

2x*()+(-3x+15)+3x*()=0

We get rid of parentheses

2x*()-3x+3x*()+15=0

We add all the numbers together, and all the variables

-3x+2x*()+3x*()+15=0

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

-3x+2x*()+3x*()=-15

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