3z+2+1=3(z+1)

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Solution for 3z+2+1=3(z+1) equation: 



3z+2+1=3(z+1)

We move all terms to the left:

3z+2+1-(3(z+1))=0

We add all the numbers together, and all the variables

3z-(3(z+1))+3=0



We calculate terms in parentheses: -(3(z+1)), so:

3(z+1)

We multiply parentheses

3z+3

Back to the equation:

-(3z+3)



We get rid of parentheses

3z-3z-3+3=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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