9=(1/5)z+12

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Solution for 9=(1/5)z+12 equation: 



9=(1/5)z+12

We move all terms to the left:

9-((1/5)z+12)=0



Domain of the equation: 5)z+12)!=0

z!=0/1

z!=0

z∈R



We add all the numbers together, and all the variables

-((+1/5)z+12)+9=0

We multiply all the terms by the denominator


-((+1+9*5)z+12)=0



We calculate terms in parentheses: -((+1+9*5)z+12), so:

(+1+9*5)z+12

We add all the numbers together, and all the variables

46z+12

Back to the equation:

-(46z+12)



We get rid of parentheses

-46z-12=0

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

-46z=12

z=12/-46

z=-6/23

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