x-(6-2x)/3=2x-4-(x+3)/2

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Solution for x-(6-2x)/3=2x-4-(x+3)/2 equation: 



x-(6-2x)/3=2x-4-(x+3)/2

We move all terms to the left:

x-(6-2x)/3-(2x-4-(x+3)/2)=0

We add all the numbers together, and all the variables

x-(-2x+6)/3-(2x-4-(x+3)/2)=0

We calculate fractions


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



We calculate terms in parentheses: +(-(2x-4-(x+3)*3)/(), so:

-(2x-4-(x+3)*3)/(

We multiply all the terms by the denominator


-(2x-4-(x+3)*3)



We calculate terms in parentheses: -(2x-4-(x+3)*3), so:

2x-4-(x+3)*3

determiningTheFunctionDomain
2x-(x+3)*3-4

We multiply parentheses

2x-3x-9-4

We add all the numbers together, and all the variables

-1x-13

Back to the equation:

-(-1x-13)



We get rid of parentheses

1x+13

We add all the numbers together, and all the variables

x+13

Back to the equation:

+(x+13)



We get rid of parentheses

x+2x/()+x+13=0

We multiply all the terms by the denominator


x*()+2x+x*()+13*()=0

We add all the numbers together, and all the variables

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

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