Algebra Exponent Problem Analysis

This problem requires us to calculate the value of a specific algebraic expression involving fractions and exponents. We are given the ratio $\frac{x}{y}$ and need to find the value of $\left(\frac{x}{y} + \frac{y}{x}\right)^{-1} \times \left(\frac{y}{x}\right)^{-1}$.

Ratio Calculation $\frac{x}{y}$

First, let's determine the value of $\frac{x}{y}$ using the provided information:

$$ \frac{x}{y} = \left(-\frac{1}{3}\right)^{-3} \div \left(\frac{2}{3}\right)^{-4} $$

We will evaluate each part of this expression:

• Evaluating $\left(-\frac{1}{3}\right)^{-3}$: Recall the rule for negative exponents: $ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n} $. $$ \left(-\frac{1}{3}\right)^{-3} = \left(\frac{-1}{3}\right)^{-3} = \left(\frac{3}{-1}\right)^{3} = (-3)^3 = -27 $$
• Evaluating $\left(\frac{2}{3}\right)^{-4}$: Using the same rule for negative exponents: $$ \left(\frac{2}{3}\right)^{-4} = \left(\frac{3}{2}\right)^{4} = \frac{3^4}{2^4} = \frac{81}{16} $$
• Performing the division to find $\frac{x}{y}$: Now, we divide the result of the first term by the result of the second term: $$ \frac{x}{y} = -27 \div \frac{81}{16} $$ Dividing by a fraction is the same as multiplying by its reciprocal: $$ \frac{x}{y} = -27 \times \frac{16}{81} $$ We can simplify this calculation. Since $81 = 3 \times 27$, we get: $$ \frac{x}{y} = -\frac{27 \times 16}{81} = -\frac{1 \times 16}{3} = -\frac{16}{3} $$

So, the ratio $\frac{x}{y}$ is equal to $-\frac{16}{3}$.

Reciprocal Ratio $\frac{y}{x}$

The term $\frac{y}{x}$ is the reciprocal of $\frac{x}{y}$.

$$ \frac{y}{x} = \frac{1}{\frac{x}{y}} = \frac{1}{-\frac{16}{3}} = -\frac{3}{16} $$

Target Expression Evaluation

Now we substitute the values of $\frac{x}{y}$ and $\frac{y}{x}$ into the expression $\left(\frac{x}{y} + \frac{y}{x}\right)^{-1} \times \left(\frac{y}{x}\right)^{-1}$.

Step 1: Calculate $\left(\frac{x}{y} + \frac{y}{x}\right)^{-1}$

• First, find the sum $\frac{x}{y} + \frac{y}{x}$: $$ \frac{x}{y} + \frac{y}{x} = -\frac{16}{3} + \left(-\frac{3}{16}\right) = -\frac{16}{3} - \frac{3}{16} $$
• To add these fractions, we find a common denominator, which is $3 \times 16 = 48$: $$ -\frac{16 \times 16}{48} - \frac{3 \times 3}{48} = \frac{-256}{48} - \frac{9}{48} = \frac{-256 - 9}{48} = -\frac{265}{48} $$
• Next, we find the inverse of this sum: $$ \left(\frac{x}{y} + \frac{y}{x}\right)^{-1} = \left(-\frac{265}{48}\right)^{-1} = -\frac{48}{265} $$

Step 2: Calculate $\left(\frac{y}{x}\right)^{-1}$

• Using the value $\frac{y}{x} = -\frac{3}{16}$, we find its inverse: $$ \left(\frac{y}{x}\right)^{-1} = \left(-\frac{3}{16}\right)^{-1} = -\frac{16}{3} $$

Step 3: Multiply the results from Step 1 and Step 2

• Finally, multiply the two results obtained: $$ \left(-\frac{48}{265}\right) \times \left(-\frac{16}{3}\right) $$
• The product of two negative numbers is positive: $$ \frac{48}{265} \times \frac{16}{3} $$
• Simplify the expression. We notice that $48$ is divisible by $3$ ($48 \div 3 = 16$): $$ \frac{16 \times 16}{265} = \frac{256}{265} $$

The calculated value of the expression is $\frac{256}{265}$.

Answer Verification and Options

Our detailed calculation shows that the value of the expression $\left(\frac{x}{y} + \frac{y}{x}\right)^{-1} \times \left(\frac{y}{x}\right)^{-1}$ is $\frac{256}{265}$.

The options provided are:

• Option 1: $\frac{165}{48}$
• Option 2: $\frac{155}{3}$
• Option 3: $\frac{145}{9}$
• Option 4: $\frac{265}{9}$

The correct answer text provided is $\frac{165}{48}$, which corresponds to Option 1. Please note that our derived result $\frac{256}{265}$ does not match any of the given options or the indicated correct answer. The steps shown above represent the accurate mathematical procedure for solving the problem as stated.