Problem 21 - Elementary Algebra

Question

Find each value. Assume the base is not zero. $$ \frac{26 x^{2} y^{5}}{4 x y^{2}} $$

Step-by-Step Solution

Answer

Question: Simplify the following expression: $$\frac{26 x^{2} y^{5}}{4 x y^{2}}$$ Answer: The simplified expression is $$\frac{13x y^{3}}{2}$$.

Step 1: Identify the factors in the numerator and the denominator

Observe the expression and identify the factors present in both the numerator and the denominator. In the numerator, we have the factors 26, x^2, and y^5. In the denominator, we have the factors 4, x, and y^2.

Step 2: Cancel out common factors

Using the exponent properties, cancel out common factors from both the numerator and the denominator. We have: $$ \frac{26 x^{2} y^{5}}{4 x y^{2}} = \frac{26}{4} \cdot \frac{x^{2}}{x} \cdot \frac{y^{5}}{y^{2}} $$ Now, using the properties of exponents, recall that a^(m-n) = a^(m)/a^(n), thus: $$ \frac{x^{2}}{x} = \frac{x^{2}}{x^{1}} = x^{(2-1)} = x^{1} = x $$ And, $$ \frac{y^{5}}{y^{2}} = y^{(5-2)} = y^{3} $$ So, the simplified expression becomes: $$ \frac{26}{4} \cdot x \cdot y^{3} $$

Step 3: Simplify the numeric part

Divide 26 by 4, reducing it to its simplest fraction. Since 26 and 4 share the common factor 2, divide both by 2: $$ \frac{26}{4} = \frac{13}{2} $$

Step 4: Write the final simplified expression

Combine the simplified numeric and algebraic expressions to obtain the final result. The final simplified expression is: $$ \frac{13}{2} \cdot x \cdot y^{3} = \frac{13x y^{3}}{2} $$

Key Concepts

Exponent Properties

Exponent properties, or laws of exponents, are the rules that govern the manipulation of expressions with exponents. Understanding these rules is crucial for simplifying algebraic expressions. A key rule is when dividing like bases, we subtract the exponents: \( a^m / a^n = a^{m-n} \). In the context of the exercise \(\frac{x^2}{x} = x^{2-1} = x\), this property allows us to cancel out common base factors and simplify the expression. Another important property is \(a^0 = 1\), useful for simplifying terms where the exponents cancel out completely.

When you encounter exponents in algebra, remember that these properties are designed to make otherwise complex calculations far more manageable. By applying them, algebra becomes a puzzle where each move is precise and leads to a simplified result.

Canceling Common Factors

Canceling common factors in an algebraic expression is like reducing a fraction to its simplest form. It's a process where you identify and divide out the common terms in the numerator and denominator. This method reduces the expression's complexity, making it easier to work with. In the given exercise, \(26x^2y^5/4xy^2\), we find that both numerators and denominators share \(x\) and \(y\) as common factors.

By canceling the common \(x\), we simplify \(x^2/x\) to \(x\), and for \(y\), \(y^5/y^2\) simplifies to \(y^3\). Canceling is a form of simplifying that assists in bringing an expression to its most manageable size, making the following mathematical operations easier to execute.

Simplifying Fractions

Simplifying fractions involves reducing the numerator and the denominator to their smallest possible numbers while keeping the same value of the fraction. In the problem, \(\frac{26}{4}\) is a fraction that can be reduced by identifying a common factor for both 26 and 4. Since both are divisible by 2, the fraction simplifies to \(\frac{13}{2}\).

This concept is essential because working with smaller numbers is usually easier and less prone to errors. When simplifying fractions, always look for the highest common factor to ensure that you're reducing the fraction fully in one step, whenever possible.

Elementary Algebra

Elementary algebra is the foundation of higher mathematics that involves variables and constants. It includes operations like addition, subtraction, multiplication, division, and the use of exponents as seen in this exercise. The process of simplification reduces complex equations to simpler forms, making it easier to solve for unknowns or understand the relationship between variables.

In our given problem, we use elementary algebra's simplification techniques to distill \(\frac{26 x^{2} y^{5}}{4 x y^{2}}\) to \(\frac{13xy^3}{2}\). Each of the steps we've taken relies on elementary algebra; it shows us patterns and helps us predict how to proceed with such problems. Using this foundational knowledge, students can build their proficiency in algebra and prepare themselves for more advanced mathematical concepts.