If \(\left(x+\frac{1}{yz}\right) - \left(y+\frac{1}{zx}\right) = \left(y+\frac{1}{zx}\right) - \left(z+\frac{1}{xy}\right)\) and \(x+z\neq2y\), then what is \(xyz\) equal to?
Algebraic Equation Simplification and Solution
The problem asks us to find the value of the product \(xyz\) given a specific algebraic equation and a condition.
Understanding the Given Equation
We are provided with the equation:
\( \left(x+\frac{1}{yz}\right) - \left(y+\frac{1}{zx}\right) = \left(y+\frac{1}{zx}\right) - \left(z+\frac{1}{xy}\right) \)
We are also given the condition that \(x+z \neq 2y\).
Step-by-Step Solution
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Simplify the Equation
First, let's expand and rearrange the terms in the given equation:
\( x + \frac{1}{yz} - y - \frac{1}{zx} = y + \frac{1}{zx} - z - \frac{1}{xy} \)
Now, group the variables and the fractional terms:
\( (x - y) + \left(\frac{1}{yz} - \frac{1}{zx}\right) = (y - z) + \left(\frac{1}{zx} - \frac{1}{xy}\right) \)
Combine the fractional terms by finding a common denominator:
\( \frac{1}{yz} - \frac{1}{zx} = \frac{x}{xyz} - \frac{y}{xyz} = \frac{x-y}{xyz} \)
\( \frac{1}{zx} - \frac{1}{xy} = \frac{y}{xyz} - \frac{z}{xyz} = \frac{y-z}{xyz} \)
Substitute these back into the rearranged equation:
\( (x - y) + \frac{x-y}{xyz} = (y - z) + \frac{y-z}{xyz} \)
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Rearrange Terms to Isolate Factors
Move all terms to one side of the equation:
\( (x - y) - (y - z) + \frac{x-y}{xyz} - \frac{y-z}{xyz} = 0 \)
Simplify the variable terms and the fractional terms:
\( x - 2y + z + \frac{(x-y) - (y-z)}{xyz} = 0 \)
\( x - 2y + z + \frac{x - 2y + z}{xyz} = 0 \)
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Factor the Equation
Notice that the term \((x - 2y + z)\) is common to both parts of the equation. Factor it out:
\( (x - 2y + z) \left(1 + \frac{1}{xyz}\right) = 0 \)
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Apply the Given Condition
The equation \((x - 2y + z) \left(1 + \frac{1}{xyz}\right) = 0\) implies that at least one of the factors must be zero. This leads to two possibilities:
- Possibility 1: \(x - 2y + z = 0\), which means \(x + z = 2y\).
- Possibility 2: \(1 + \frac{1}{xyz} = 0\).
We are given the condition that \(x+z \neq 2y\). This explicitly rules out Possibility 1.
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Determine the Value of xyz
Since Possibility 1 is ruled out by the condition \(x+z \neq 2y\), Possibility 2 must be true:
\( 1 + \frac{1}{xyz} = 0 \)
Now, solve for \(xyz\):
\( \frac{1}{xyz} = -1 \)
\( xyz = -1 \)
Conclusion
By simplifying the given algebraic equation and applying the condition \(x+z \neq 2y\), we find that the value of the product \(xyz\) must be \(-1\).
Important Questions from Arithmetic Progression
If the arithmetic mean of a, b, c is \(\rm \frac M 3\) and \(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c} \) , then the arithmetic mean of a 2, b 2, c 2 is
A person saves Rs. 1000 more than he did the previous year. If he saves Rs. 2000 in the first year, in how many years will he save Rs. 170000?
How many natural numbers lie between 3 and 200 which are divisible by 7?
The arithmetic mean of 16 student's scores is 320. Find the sum of these scores.
How many three-digit numbers are divisible by 5?
