Problem 101 - Elementary Algebra

Question

Write the following expressions using only positive exponents. Assume all variables are nonzero. $$ \frac{3 x^{2} y^{-2}(x-5)}{9^{-1}(x+5)^{3}} $$

Step-by-Step Solution

Answer

Question: Rewrite the given expression with only positive exponents: $$ \frac{3 x^{2} y^{-2}(x-5)}{9^{-1}(x+5)^{3}} $$ Answer: The expression with only positive exponents is: $$ \frac{27 x^2 (x-5)}{y^2 (x+5)^3} $$

Step 1: Identify terms with negative exponents

The given expression is: $$ \frac{3 x^{2} y^{-2}(x-5)}{9^{-1}(x+5)^{3}} $$ We can see that there are two terms with negative exponents: \(y^{-2}\) and \(9^{-1}\).

Step 2: Apply the reciprocal exponent rule

We apply the rule \(a^{-n} = \frac{1}{a^n}\) to both terms with negative exponents: $$ \frac{3 x^{2} \frac{1}{y^{2}}(x-5)}{\frac{1}{9}(x+5)^{3}} $$

Step 3: Simplify the expression

Now, we can simplify the expression: $$ \begin{aligned} \frac{3 x^{2} \frac{1}{y^{2}}(x-5)}{\frac{1}{9}(x+5)^{3}} &= \frac{3 \cdot 9 x^{2}(x-5)}{y^2(x+5)^3} \\ &= \frac{27 x^2 (x-5)}{y^2 (x+5)^3} \end{aligned} $$ The final expression with only positive exponents is: $$ \frac{27 x^2 (x-5)}{y^2 (x+5)^3} $$

Key Concepts

negative exponents

Negative exponents can be a bit tricky at first, but they're actually quite straightforward when you get the hang of them. When you see a negative exponent, it means you're dealing with division in a clever way. Essentially, a negative exponent like \(a^{-n}\) represents the reciprocal of the base raised to the corresponding positive exponent: \(a^{-n} = \frac{1}{a^n}\).
This concept allows us to express quantities in alternate forms without changing their actual values. For instance, with \(y^{-2}\), it means \(\frac{1}{y^2}\). Negative exponents are great tools in algebra because they help simplify complex expressions and equations, making calculations much more manageable.

reciprocal exponent rule

The reciprocal exponent rule is a fundamental principle in algebra that helps us handle negative exponents. According to this rule, if you have a term with a negative exponent, such as \(a^{-n}\), you can convert it to a positive exponent by taking the reciprocal: \(a^{-n} = \frac{1}{a^n}\).
This rule is handy because it turns division into multiplication, which is often easier to work with in equations and expressions.
Let's see it in action:

• For the term \(9^{-1}\), applying the rule gives us \(\frac{1}{9}\).
• For \(y^{-2}\), we transform it into \(\frac{1}{y^2}\).

Using the reciprocal exponent rule keeps equations tidy and expressions efficient.

simplification

Simplification in algebra involves reducing expressions to their simplest form, making them easier to understand and work with. This often means eliminating negative exponents, combining like terms, and performing basic arithmetic operations.
In the given exercise, once we've applied the reciprocal exponent rule, we end up with intermediate steps that need to be simplified further.
For instance:

• Replace negative terms using their reciprocal forms, like converting \(\frac{1}{9}\) by multiplying instead of dividing to eliminate fractions within fractions, resulting in \(9\) in the numerator.
• Multiply the constants and coefficients, such as \(3 \times 9\) transforming to \(27\).

Simplifying ensures that the expression \(\frac{27x^2 (x-5)}{y^2 (x+5)^3}\) is in its most efficient and manageable form.

algebraic expressions

Algebraic expressions are combinations of variables, numbers, and operations (like addition, subtraction, multiplication, and division). They provide a powerful way to represent real-world situations through mathematics. An expression can include constants, coefficients, variables, and exponents, just like the problem \(\frac{3 x^{2} y^{-2}(x-5)}{9^{-1}(x+5)^{3}}\).
Here are a few key components to understand:

• Variables: Symbols that represent unknown values (e.g., \(x\), \(y\)).
• Constants and Coefficients: Fixed values and multipliers of variables (e.g., 3, 9).
• Operators: Indicate the operations, like \(+\), \(-\), \(\times\), and \(\div\).
• Terms: Parts of the expression, which could be a single number, a variable, or a product of both.

Understanding how these components work together is the foundation of tackling more complex algebraic scenarios, enabling simplifications and transformations into results like \(\frac{27x^2 (x-5)}{y^2 (x+5)^3}\).