+0  
 
+1
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avatar+1050 

Let x, y, and z be nonzero real numbers. Find all possible values of x/ | x | + y/ | y | + z/ | z | + xyz/ | xyz |.  List your values in increasing order, separated by commas.

 

Thanks so much!

 

Edit: I solved this one too, it's -4,0,4

AnonymousConfusedGuy  Dec 11, 2017
edited by AnonymousConfusedGuy  Dec 11, 2017

Best Answer 

 #1
avatar+7056 
+1

I know you solved this already, but I wanted to try it too.

 

If  x ,  y , and  z  are positive...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{x}{x}+\frac{y}{y}+\frac{z}{z}+\frac{xyz}{xyz}\,=\,1+1+1+1\,=\,4\)

 

If only  x  is negative...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{y}{y}+\frac{z}{z}+\frac{-|x|yz}{|xyz|}\,=\,-1+1+1-1\,=\,0\)

 

If only  y  or only  z  is negative, the result will also be  0 .

 

If  x  and  y  are negative, and  z  is positive...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{z}{z}+\frac{(-|x|)(-|y|)z}{|xyz|}\,=\,-1-1+1+1\,=\,0\)

 

If any two are negative, the result will also be  0  .

 

If  x ,  y , and  z  are negative...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{-|z|}{|z|}+\frac{(-|x|)(-|y|)(-|z|)}{|xyz|}\,=\,-1-1-1-1\,=\,-4\)

 

smileysmiley

hectictar  Dec 11, 2017
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2+0 Answers

 #1
avatar+7056 
+1
Best Answer

I know you solved this already, but I wanted to try it too.

 

If  x ,  y , and  z  are positive...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{x}{x}+\frac{y}{y}+\frac{z}{z}+\frac{xyz}{xyz}\,=\,1+1+1+1\,=\,4\)

 

If only  x  is negative...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{y}{y}+\frac{z}{z}+\frac{-|x|yz}{|xyz|}\,=\,-1+1+1-1\,=\,0\)

 

If only  y  or only  z  is negative, the result will also be  0 .

 

If  x  and  y  are negative, and  z  is positive...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{z}{z}+\frac{(-|x|)(-|y|)z}{|xyz|}\,=\,-1-1+1+1\,=\,0\)

 

If any two are negative, the result will also be  0  .

 

If  x ,  y , and  z  are negative...

 

\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{-|z|}{|z|}+\frac{(-|x|)(-|y|)(-|z|)}{|xyz|}\,=\,-1-1-1-1\,=\,-4\)

 

smileysmiley

hectictar  Dec 11, 2017
 #2
avatar+86649 
+1

Nice, hectictar!!!!!....I like this one    ....

 

 

 

cool cool cool

CPhill  Dec 11, 2017

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