(2x+1)/3+(3x+7)/5=(x+14)/5

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Solution for (2x+1)/3+(3x+7)/5=(x+14)/5 equation: 



(2x+1)/3+(3x+7)/5=(x+14)/5

We move all terms to the left:

(2x+1)/3+(3x+7)/5-((x+14)/5)=0

We calculate fractions


2x/()+(9x+21)/()+(-((x+14)*3)/()=0



We calculate terms in parentheses: +(-((x+14)*3)/(), so:

-((x+14)*3)/(

We multiply all the terms by the denominator


-((x+14)*3)



We calculate terms in parentheses: -((x+14)*3), so:

(x+14)*3

We multiply parentheses

3x+42

Back to the equation:

-(3x+42)



We get rid of parentheses

-3x-42

Back to the equation:

+(-3x-42)



We get rid of parentheses

2x/()+(9x+21)/()-3x-42=0

We multiply all the terms by the denominator


2x+(9x+21)-3x*()-42*()=0

We add all the numbers together, and all the variables

2x+(9x+21)-3x*()=0

We get rid of parentheses

2x+9x-3x*()+21=0

We add all the numbers together, and all the variables

11x-3x*()+21=0

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

11x-3x*()=-21

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