3x-(2x+6)/3=16-(x+2)/2

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Solution for 3x-(2x+6)/3=16-(x+2)/2 equation: 



3x-(2x+6)/3=16-(x+2)/2

We move all terms to the left:

3x-(2x+6)/3-(16-(x+2)/2)=0

We calculate fractions


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



We calculate terms in parentheses: +(-(16-(x+2)*3)/(), so:

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

We multiply all the terms by the denominator


-(16-(x+2)*3)



We calculate terms in parentheses: -(16-(x+2)*3), so:

16-(x+2)*3

determiningTheFunctionDomain
-(x+2)*3+16

We multiply parentheses

-3x-6+16

We add all the numbers together, and all the variables

-3x+10

Back to the equation:

-(-3x+10)



We get rid of parentheses

3x-10

Back to the equation:

+(3x-10)



We get rid of parentheses

3x+(-2x)/()+3x-10=0

We multiply all the terms by the denominator


3x*()+(-2x)+3x*()-10*()=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

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