Question

If \(x, y, z\) are three distinct positive real numbers sucn and \(x+y+z=1\), then the value of \(\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\) is: (a) 16 (b) 8 (c) 4 (d) 1

Short answer

Based on the step-by-step solution above, the value of \(\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\) for the given example is \(7\). However, none of the given options match this value. Please check the problem statement and the provided options to ensure their correctness. If not, it is possible that the problem is flawed.

Step-by-step solution

Calculate the new expressions

We have the given expression \(\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\). We can rewrite the terms inside the brackets as follow: \(\frac{1}{x}-1 = \frac{1-x}{x}\), \(\frac{1}{y}-1 = \frac{1-y}{y}\), \(\frac{1}{z}-1 = \frac{1-z}{z}\). Now, substitute these new expressions back into the original expression.

Substitute the new expressions

We substitute the new expressions back into the original expression \(\left(\frac{1-x}{x}\right)\left(\frac{1-y}{y}\right)\left(\frac{1-z}{z}\right)\).

Multiply the expressions

Now we need to multiply the three expressions: \(\frac{(1-x)(1-y)(1-z)}{xyz}\).

Use \(x + y + z = 1\) to substitute

Since \(x + y + z = 1\), we have \(z = 1 - x - y\). Substitute \(z = 1 - x - y\) into the expression: \(\frac{(1-x)(1-y)(1-(1-x-y))}{xy(1-x-y)}\).

Simplify the expression

Now simplify the expression: \(\frac{(1-x)(1-y)(x+y)}{xy(1-x-y)}\). Note that we have \((1-x-y)\) both in the numerator and denominator, so we can cancel them: \(\frac{(1-x)(1-y)}{xy}\).

Expand the expression

Expand the expression: \(\frac{1-x-x(y-1)}{xy}=\frac{1-x(2-y)}{xy}\). We need to find a numerical value for this expression. Since \(x, y, z\) are distinct positive real numbers with \(x+y+z=1\), let's take \(x=\frac{1}{3}\), \(y=\frac{1}{6}\), and \(z=\frac{1}{2}\) as an example.

Substitute the values of x and y

Substitute the values of \(x\) and \(y\) into the expression: \(\frac{1-\frac{1}{3}(2-\frac{1}{6})}{\frac{1}{3}\times\frac{1}{6}}\).

Simplify and compute the value

Now simplify and compute the value: \(\frac{1-\frac{1}{3}(\frac{11}{6})}{\frac{1}{18}} = \frac{1-\frac{11}{18}}{\frac{1}{18}}= \frac{\frac{7}{18}}{\frac{1}{18}}=7\). Now, we see that none of the given options match the computed value of \(7\). There might be an error in the given options. Double-check the problem statement and the options to make sure they are correct.

Key concepts

Positive Real Numbers

Positive real numbers are numbers greater than zero on the number line. They include all fractions, integers, and decimals greater than zero. A positive real number can be expressed in the form of a fraction, such as \(\frac{1}{2}\) or a whole number like \(4\), and they are part of the larger set known as real numbers, which also include negative numbers and zero.

• These numbers do not include imaginary, complex, or negative numbers.
• Any expression derived from positive real numbers will also yield a positive result if only addition, multiplication, or exponentiation with even power is involved.

Understanding positive real numbers is critical when working with algebraic expressions that involve ratios or summations, as in the given exercise, where \(x + y + z = 1\) signifies that combined, these positive values sum to one.

Simplification

Simplification in algebra is the process of altering the form of an expression to make it easier to understand or work with, without changing its value. In the provided exercise, simplification involves a series of steps to translate a complex expression into a more manageable form.

• Initially, terms such as \(\frac{1}{x} - 1\) were rewritten as \(\frac{1-x}{x}\), breaking them down into simpler fractions.
• Simplifying expressions often involves canceling like terms or reducing fractions to their lowest terms.

The simplification in this problem was essential to understanding and calculating the complex product of three expressions. By simplifying, we were able to work through tedious algebraic manipulations, ultimately finding the expression's value through a series of logical and clear transformations.

Distinct Values

In algebra, distinct values mean that each variable represents a different number. In this context, \(x\), \(y\), and \(z\) are noted as distinct since no two variables have the same value. This condition is pivotal when solving expressions that depend on variations among variables.

• This distinction prevents the expressions from simplifying to trivially zero or some other simplified form shared between commonplace values.
• The distinct nature is mathematically established to avoid infinite solutions or redundant computations.

For such problems, assuming different rational values for these variables, like \(x=\frac{1}{3}\), \(y=\frac{1}{6}\), and \(z=\frac{1}{2}\), ensures practical calculations while adhering to \(x + y + z = 1\).

Expression Expansion

Expression expansion involves distributing the terms to break down and simplify the algebraic expressions further. It lays out each individual element for easier manipulation and simplification.

• The exercise included expanding the expression \(\frac{(1-x)(1-y)(x+y)}{xy(1-x-y)}\).
• Expanding helps to cancel out like terms as seen with the \(1-x-y\) term, simplifying and reducing the complex expressions into more straightforward arithmetic or numerical calculations.

By expanding and carefully computing each step, algebraic expressions reveal their inherent patterns, making the process of finding solutions more manageable and systematic.