41=1+4/5z

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Solution for 41=1+4/5z equation: 



41=1+4/5z

We move all terms to the left:

41-(1+4/5z)=0



Domain of the equation: 5z)!=0

z!=0/1

z!=0

z∈R



We add all the numbers together, and all the variables

-(4/5z+1)+41=0

We get rid of parentheses

-4/5z-1+41=0

We multiply all the terms by the denominator


-1*5z+41*5z-4=0

Wy multiply elements

-5z+205z-4=0

We add all the numbers together, and all the variables

200z-4=0

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

200z=4

z=4/200

z=1/50

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