Problem 24 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Find the product below by multiplying the numerators and denominators, then simplifying. Then fi nd the product by simplifying each expression, then multiplying. Which method do you prefer? Explain. $$ \frac{4 x^2 y}{2 x^3} \cdot \frac{12 y^4}{24 x^2} $$

Step-by-Step Solution

Answer

The simplified expression of the given fraction is \( \frac{y^5}{x^3} \). The preferable method depends on personal comfort and habit. Both methods give the same final result.

Step 1: Multiply Numerators and Denominators

In this step, multiply the numerators together to get the new numerator,and the denominators together to get the new denominator. Here, the expression becomes \( \frac{4x^2y \cdot 12y^4}{2x^3 \cdot 24x^2} = \frac{48x^2y^5}{48x^5} \)

Step 2: Simplify Resulting Fraction

Now, simplify the resulting fraction. The numbers in both the numerator and denominator simplify to 1, and the variable terms simplify to \(x^{-3} y^{5}\). This gives a simplified fraction: \(1x^{-3} y^{5}\) or simply \( \frac{y^5}{x^3} \).

Step 3: Simplify Before Multiplication

The second part of the problem requires simplification before multiplication. Simplify each expression before multiplying, \( \frac{4 x^2 y}{2 x^3} = \frac{2y}{x} \) and \( \frac{12 y^4}{24 x^2} = \frac{y^{2}}{2x^{2}} \).

Step 4: Multiply Simplified Expressions

Next, multiply the simplified expressions from step 3 and obtain \( \frac{2y \cdot y^2}{x \cdot 2x^2} = \frac{2y^3}{2x^3} \) which simplifies further to \( \frac{y^{3}}{x^{3}} \).

Step 5: Compare Methods

Finally, both methods give the same result. The preferable method depends on personal comfort and habit. Some might find it easier to multiply first and then simplify, but others might find it more straightforward to simplify before multiplication.

Key Concepts

Multiplying Fractions

When multiplying fractions, you need to multiply the numerators (the top parts of the fractions) to get a new numerator, and you multiply the denominators (the bottom parts of the fractions) to get a new denominator.
This is a straightforward approach that leverages basic multiplication skills.

• For example, to multiply \( \frac{4x^2y}{2x^3} \) and \( \frac{12y^4}{24x^2} \), you first multiply the numerators: \( 4x^2y \cdot 12y^4 = 48x^2y^5 \).
• Next, multiply the denominators: \( 2x^3 \cdot 24x^2 = 48x^5 \).

The new fraction becomes \( \frac{48x^2y^5}{48x^5} \).
However, this fraction is not in its simplest form, and thus would need to be simplified further, as shown in the next section.
This method is usually preferred in cases where simplifying before multiplying does not notably reduce the complexity.

Simplifying Expressions

Simplifying expressions involves breaking down each part of the fraction to its simplest form before performing any operations like multiplication or division.
This can make calculations easier and reduce the chances of errors due to large or complex numbers.

• In our example, instead of multiplying right away, simplify \( \frac{4x^2y}{2x^3} \) to \( \frac{2y}{x} \).
• Similarly, simplify \( \frac{12y^4}{24x^2} \) to \( \frac{y^2}{2x^2} \).

Now you multiply these simplified forms: \( \frac{2y}{x} \times \frac{y^2}{2x^2} = \frac{2y^3}{2x^3} \).
This simplifies once more to \( \frac{y^3}{x^3} \).
Simplifying before performing arithmetic can often lead to a clearer solution path, but requires careful attention to ensure each part is reduced correctly.

Variables and Exponents

Understanding variables and exponents is crucial in algebra and especially when simplifying expressions.

• Variables represent unknown values and are often denoted by letters like \( x \) or \( y \).
• Exponents tell you how many times to multiply the base by itself. For example, \( x^3 \) means \( x \times x \times x \).

In the problem \( \frac{4x^2y}{2x^3} \cdot \frac{12y^4}{24x^2} \), exponents allow you to apply the laws of powers, such as \( x^a \cdot x^b = x^{a+b} \) to simplify expressions.

• Look at the fraction \( \frac{48x^2y^5}{48x^5} \), here the common base \( x \) is in both the numerator and denominator, allowing you to simplify using the rule \( x^a/x^b = x^{a-b} \).
• This leaves you with \( x^{2-5} = x^{-3} \) indicating that \( x \) should be in the denominator as \( x^3 \).

Thus, variables and exponents are the backbone to manipulating and simplifying algebraic expressions efficiently.