(3(7y+1)/3)+2=7y+3

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Solution for (3(7y+1)/3)+2=7y+3 equation: 



(3(7y+1)/3)+2=7y+3

We move all terms to the left:

(3(7y+1)/3)+2-(7y+3)=0

We get rid of parentheses

(3(7y+1)/3)-7y-3+2=0

We multiply all the terms by the denominator


(3(7y+1)-7y*3)-3*3)+2*3)=0



We calculate terms in parentheses: +(3(7y+1)-7y*3), so:

3(7y+1)-7y*3

We multiply parentheses

21y-7y*3+3

Wy multiply elements

21y-21y+3

We add all the numbers together, and all the variables

3

Back to the equation:

+(3)



We add all the numbers together, and all the variables

=0

y=0/1

y=0

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