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2+1/{x+1/(y+1/z)}=37/13 Find the values of x,y and z.(numerical values)Please solve...

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madmoni06 | Student | Honors

Posted February 6, 2013 at 3:28 PM via web

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2+1/{x+1/(y+1/z)}=37/13

Find the values of x,y and z.(numerical values)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 6, 2013 at 4:20 PM (Answer #1)

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Since the problem provides only one equation and three unknowns, x,y,z, then you cannot find the numerical values of each unknown, because the number of variables overcome the number of given equation.

You can find the numerical values of x,y,z only if the problem provides two more relations that involve x,y and z.

The only thing you can do is to find an equation that relates the variables x,y and z such that:

`2 + 1/(x + 1/(y + 1/z)) = 37/13`

`2 + 1/(x + z/(yz + 1)) = 37/13`

`2 + (yz + 1)/(xyz + x + z) = 37/13`

`(2xyz + 2x + 2z + yz + 1)/(xyz + x + z) = 37/13`

`37xyz + 37x + 37z = 26xyz + 26x + 26z + 13yz + 13`

`11xyz + 11x + 11z = 13yz + 13`

`11(xyz + x + z) = 13(yz + 1)`

Hence, evaluating the equation that relates the variables x,y and z yields `11(xyz + x + z) = 13(yz + 1).`

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