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If \(x+\frac{1}{y}=1\) and \(y+\frac{1}{z}=1\), what is the value of the product \(xyz\)?

 Jan 19, 2021
 #1
avatar+536 
-1

0

 

x=0, y=1, z=0

 Jan 19, 2021
 #2
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0

z cannot equal zero......   1/z    in second equation

 Jan 19, 2021
 #3
avatar+310 
+2

Multiply both sides of the first equation by \(y \) and both sides of the second equation by  to obtain \(z\)

\(\begin{align*} xy+1 &= y \\ yz+1 &= z. \end{align*}\)

Substituting \(xy+1\) for \(y\) in the second equation, we find

\((xy+1)z+1=z,\)

which simplifies to

\(xyz+z+1=z.\)

Subtracting \(z+1\) from both sides, we find that \(xyz=z-(z+1)=\boxed{-1}.\)

 Jan 19, 2021
 #4
avatar+116126 
+2

x + 1/y  =  1               y +  1/z  =  1

 

(xy + 1) / y  = 1        (yz  + 1) /  z  = 1

 

xy +  1 =  y              yz +  1  = z

 

xy =  y -1                 1 = z - zy

 

x = (y-1)/y                1 = z ( 1 - y)

 

                                 z  = 1/(1-y)

 

So

 

xyz  =    (y-1) / y  *   y  *  1/(1-y)       =    (y-1)  /  (1-y)  =   -(1-y)  /( 1-y)  =    -1 

 

 

cool cool cool

 Jan 19, 2021

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