3(z+4)+z=4(z+3)

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Solution for 3(z+4)+z=4(z+3) equation: 



3(z+4)+z=4(z+3)

We move all terms to the left:

3(z+4)+z-(4(z+3))=0

We add all the numbers together, and all the variables

z+3(z+4)-(4(z+3))=0

We multiply parentheses

z+3z-(4(z+3))+12=0



We calculate terms in parentheses: -(4(z+3)), so:

4(z+3)

We multiply parentheses

4z+12

Back to the equation:

-(4z+12)



We add all the numbers together, and all the variables

4z-(4z+12)+12=0

We get rid of parentheses

4z-4z-12+12=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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