(x+2)/3=2-(x+2)/7

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Solution for (x+2)/3=2-(x+2)/7 equation: 



(x+2)/3=2-(x+2)/7

We move all terms to the left:

(x+2)/3-(2-(x+2)/7)=0

We calculate fractions


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



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

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

We multiply all the terms by the denominator


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



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

2-(x+2)*3

determiningTheFunctionDomain
-(x+2)*3+2

We multiply parentheses

-3x-6+2

We add all the numbers together, and all the variables

-3x-4

Back to the equation:

-(-3x-4)



We get rid of parentheses

3x+4

Back to the equation:

+(3x+4)



We get rid of parentheses

x/()+3x+4=0

We multiply all the terms by the denominator


x+3x*()+4*()=0

We add all the numbers together, and all the variables

x+3x*()=0

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