1/(z+2)-1/2=z/(z+2)

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Solution for 1/(z+2)-1/2=z/(z+2) equation: 



1/(z+2)-1/2=z/(z+2)

We move all terms to the left:

1/(z+2)-1/2-(z/(z+2))=0



Domain of the equation: (z+2)!=0

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

z!=-2

z∈R





Domain of the equation: (z+2))!=0

z∈R



We calculate fractions


2z/((z+2)*(z+2))*2)+(-(z*(z+2)*2)/((z+2)*(z+2))*2)+(-1*(z+2)*(z+2)))/((z+2)*(z+2))*2)=0

We add all the numbers together, and all the variables

2z/((z+2)*(z+2))*2)+(-(z*(z+2)*2)/((z+2)*(z+2))*2)+(-1*(z+2)*(z+2)))/((z+2)*(z=0

We calculate fractions


2)*(z+(2z-(z*(z+2)*2)*(z/(((z+2)*(z+2))*2)+(*(z+(-1*(z+2)*(z+2)))*((z+2)*z/(((z+2)*(z+2))*2)+(*(z=0

We can not solve this equation

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