Question

if x,y, and z are positive numbers satisfying x+1/y=4, y+1/z=1, and z+1/x=7/3 then value of xyz is
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Answer 1

Answer:

Given:

x+1/y=4

y+1/z=1

z+1/x=7/3

1/y=4-x

x+1/4-x=4

x=3/2

1/z=1-y

y+1/y-1=1

y=2/5

1/x=7/3-z

z+1/7/3-z=7/3

z=5/3

xyz=3/2×2/5×5/3

xyz=1

Answer 2

Given,

$x+\frac{1}{y} = 4$

$y+\frac{1}{z} =1$

$z + \frac{1}{x} = \frac{7}{3}$

To find,

The value of xyz.

Solution,

The value of xyz is 1.

We can simply solve this mathematical problem by the following process.

We know that,

⇒ $x + \frac{1}{y} = 4$  (Equation 1)

⇒ $y +\frac{1}{z} = 1$   (Equation 2)

⇒ $z+ \frac{1}{x} = \frac{7}{3}$   (Equation 3)

Multiplying the above three equations,

⇒ $(x+\frac{1}{y} )(y + \frac{1}{z})(z+\frac{1}{x}) = 4* 1*\frac{7}3}$

⇒ $xyz + \frac{xy}{y}+\frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} +\frac{z}{xz}+\frac{1}{xyz} = \frac{28}{3}$

⇒ $xyz + x+y+z+\frac{1}{x}+\frac{1}{y} +\frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}$

⇒ $xyz + 4 +1+\frac{7}{3} + \frac{1}{xyz}=\frac{28}{3}$

⇒ $xyz + \frac{1}{xyz} + \frac{22}{3} = \frac{28}{3}$

⇒ $xyz + \frac{1}{xyz} = \frac{6}{3}$

Let us take xyz = t

⇒ $t + \frac{1}{t} = 2$

⇒ t²+ 1 = 2t

⇒ t² - 2t + 1 = 0

⇒ t² - t - t + 1 = 0

⇒ t (t -1) -1 (t-1) = 0

⇒ (t-1)(t-1) = 0

⇒ t = 1

⇒ xyz = 1

As a result, the value of xyz is 1.