3(z+1)=4(6-2)

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Solution for 3(z+1)=4(6-2) equation: 



3(z+1)=4(6-2)

We move all terms to the left:

3(z+1)-(4(6-2))=0

We add all the numbers together, and all the variables

3(z+1)-(44)=0

We add all the numbers together, and all the variables

3(z+1)-44=0

We multiply parentheses

3z+3-44=0

We add all the numbers together, and all the variables

3z-41=0

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

3z=41

z=41/3

z=13+2/3

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