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(3x-2)!+=1

Guest Aug 15, 2017
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Let's think about this problem. There are 2 solutions to \(y!=1\). It is 0 and 1. Therefore, we need to set \(3x-2\) equal to both 0 and 1:

 

\(3x-2=0\) Add 2 to both sides.
\(3x=2\) Divide by 3 on both sides.
\(x=\frac{2}{3}\)  
   

 

Let's do the other one now!

 

\(3x-2=1\) Add 2 to both sides.
\(3x=3\) Divide by 3 on both sides.
\(x=1\)  
   

 

Therefore, x=1 and 2/3

TheXSquaredFactor  Aug 15, 2017

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