(1)/(2)(4z+2)=16

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Solution for (1)/(2)(4z+2)=16 equation: 



(1)/(2)(4z+2)=16

We move all terms to the left:

(1)/(2)(4z+2)-(16)=0



Domain of the equation: 2(4z+2)!=0

z∈R



We multiply all the terms by the denominator


-16*2(4z+2)+1=0

Wy multiply elements

-32z(4+1=0

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