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If x + y + z = 26 and 1/x + 1/y + 1/z = 31, then find x/y + y/x + x/z + z/x + y/z + z/y.

 Jun 18, 2020
 #1
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If

\(x + y + z = 26\) and \(\dfrac1x + \dfrac1y + \dfrac1z = 31\), then find \(\dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy\).

 

\(\begin{array}{|rcll|} \hline && \dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy \\\\ &=& \dfrac xy +\dfrac xz + \dfrac yx + \dfrac yz + \dfrac zx + \dfrac zy \\\\ &=& x\left(\dfrac 1y +\dfrac 1z \right) + y\left(\dfrac 1x + \dfrac 1z\right) + z\left(\dfrac 1x + \dfrac 1y\right) \\\\ && \boxed{\dfrac1x + \dfrac1y + \dfrac1z = 31\\ \dfrac1x + \dfrac1y = 31- \dfrac1z \\ \dfrac1x + \dfrac1z = 31- \dfrac1y \\ \dfrac1y + \dfrac1z = 31-\dfrac1x } \\\\ &=& x\left(31-\dfrac1x\right) + y\left(31- \dfrac1y\right) + z\left(31- \dfrac1z\right) \\\\ &=& 31x-1 +31y-1 + 31z-1 \\\\ &=& 31(x+y+z)-3\quad | \quad \mathbf{x + y + z = 26} \\ &=& \mathbf{803} \\ \hline \end{array}\)

 

 

laugh

 Jun 19, 2020

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