z+4z+12-22=3z+32

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Solution for z+4z+12-22=3z+32 equation: 



z+4z+12-22=3z+32

We move all terms to the left:

z+4z+12-22-(3z+32)=0

We add all the numbers together, and all the variables

5z-(3z+32)-10=0

We get rid of parentheses

5z-3z-32-10=0

We add all the numbers together, and all the variables

2z-42=0

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

2z=42

z=42/2

z=21

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