(2)/(3)z+1=13

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Solution for (2)/(3)z+1=13 equation: 



(2)/(3)z+1=13

We move all terms to the left:

(2)/(3)z+1-(13)=0



Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



We add all the numbers together, and all the variables

2/3z-12=0

We multiply all the terms by the denominator


-12*3z+2=0

Wy multiply elements

-36z+2=0

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

-36z=-2

z=-2/-36

z=1/18

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