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# HELP!!!!!

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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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