Problem 35 - Beginning and Intermediate Algebra

Question

Simplify. $$\frac{x^{2} y^{-2}}{x^{-1} y}$$

Step-by-Step Solution

Answer

The simplified expression is \(x^3 y^{-3}\).

Step 1: Identify exponent rules required to simplify

We have the expression: \(\frac{x^{2} y^{-2}}{x^{-1} y}\). We need to apply the following exponent rules: 1. When dividing with the same base, we subtract the exponents: \(\frac{a^n}{a^m} = a^{n-m}\) 2. When multiplying with the same base, we add the exponents: \(\frac{a^n}{a^m} = a^{n+m}\)

Step 2: Apply the exponent rules to x

First, let's focus on the terms with the base x. We have \(x^2\) in the numerator and \(x^{-1}\) in the denominator. Using rule 1, when dividing with the same base: \[\frac{x^2}{x^{-1}} = x^{2 - (-1)}\] This simplifies to: \[x^{2 + 1} = x^3\]

Step 3: Apply the exponent rules to y

Now, let's focus on the terms with the base y, which are \(y^{-2}\) in the numerator and y in the denominator. Let's first rewrite \(y\) in the denominator as an equivalent expression with an exponent. Since any number raised to the power of 1 is equal to itself, we can rewrite \(y\) as \(y^1\). Now, using rule 1, when dividing with the same base: \[\frac{y^{-2}}{y^1}= y^{-2-(1)}\] This simplifies to: \[y^{-2 - 1} = y^{-3}\]

Step 4: Combine the simplified expressions of x and y

Now that we have simplified the expressions for x and y, we can combine them: \[\frac{x^{2} y^{-2}}{x^{-1} y} = x^3 y^{-3}\] So the simplified expression is: \[x^3 y^{-3}\]

Key Concepts

Exponent Rules

Exponent rules are essential when working with expressions that involve powers. They provide a systematic way to simplify expressions by applying certain rules related to the powers of numbers and variables.
Here are a few fundamental rules you should know:

• Product of Powers Rule: When multiplying the same base, add the exponents. For example, if you have \(a^m \times a^n\), it becomes \(a^{m+n}\).
• Quotient of Powers Rule: When dividing the same base, subtract the exponents. For example, \(\frac{a^m}{a^n} = a^{m-n}\).
• Power of a Power Rule: When raising a power to another power, multiply the exponents, such as \((a^m)^n = a^{m \times n}\).

Understanding these rules allows you to navigate and simplify complex exponent expressions efficiently. They are particularly useful in algebra, a crucial component of high school and college mathematics.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form while maintaining their equivalence. This process makes expressions easier to work with, especially when solving equations or comparing expressions.
To do this, you typically:

• Apply mathematical operations, such as addition, subtraction, multiplication, and division.
• Use exponent rules to manage powers appropriately.
• Cancel out terms when possible to decrease the number of components in an expression.

In our problem, simplifying the expression \( \frac{x^2 y^{-2}}{x^{-1} y} \) involves applying the quotient of powers rule to both \(x\) and \(y\). For \(x\), this results in \(x^{2-(-1)}\), simplifying to \(x^3\). For \(y\), apply the rule to get \(y^{-2-1}\), which simplifies to \(y^{-3}\).
Once simplified, the expression becomes \(x^3 y^{-3}\), which is significantly easier to interpret and use in further calculations.

Dividing Exponents

When dividing exponents, especially those with variables, it's important to follow the rules closely to avoid errors. The quotient of powers rule is key. Here, divide by subtracting the exponent in the denominator from the exponent in the numerator, as shown in the general rule \(\frac{a^m}{a^n} = a^{m-n}\).
In our original exercise, we divided the exponents for both variables, \(x\) and \(y\).

• For \(x\): After dividing \(x^2\) by \(x^{-1}\), we subtract: \(2 - (-1) = 3\), resulting in \(x^3\).
• For \(y\): After dividing \(y^{-2}\) by \(y^1\), we subtract: \(-2 - 1 = -3\), resulting in \(y^{-3}\).

The expression \(\frac{x^{2} y^{-2}}{x^{-1} y} \) ultimately simplifies to \(x^3 y^{-3} \).
Understanding dividing exponents is vital as it helps handle more complex algebraic expressions effectively and is foundational knowledge in higher mathematics.