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
How many two-digit numbers are divisible by 3 ?
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?
A car starts with a speed of 60 km/h with its speed increasing every one hour by 5 km/h. In how many hours will it cover 435 kms?
How many natural numbers lie between 3 and 200 which are divisible by 7?
