Problem 12 - Elementary and Intermediate Algebra

Question

The type of multiplication that is used to build rational expressions is shown below. Fill in the blanks. a. \(\frac{4 x}{5} \cdot \frac{2}{2}=\frac{\underline{\phantom{xx}}}{10}\) b. \(\frac{3}{t} \cdot \frac{t-2}{t-2}=\frac{\underline{\phantom{xx}}}{t(t-2)}\) c. \(\frac{m+1}{m-3} \cdot \frac{m-5}{m-5}=\frac{\underline{\phantom{xx}}}{(m-3)(m-5)}\)

Step-by-Step Solution

Answer

a. 8x b. 3 c. m+1

Step 1: Multiplying Numerator and Denominator

For (a), we are given \( \frac{4x}{5} \cdot \frac{2}{2} \). Multiply the numerators: \( 4x \cdot 2 = 8x \). The denominator is \( 5 \cdot 2 = 10 \). Thus, the complete expression is \( \frac{8x}{10} \).

Step 2: Simplifying Expression a

The fraction for part (a) becomes \( \frac{8x}{10} \). The missing numerator from the blank is \( 8x \).

Step 3: Using Identity Property

For (b), \( \frac{3}{t} \cdot \frac{t-2}{t-2} \) shows the identity property of 1, as \( \frac{t-2}{t-2} = 1 \). Thus, the product is \( \frac{3}{t} \). Here, \( 3 \) is the numerator, and \( t \cdot (t-2) = t(t-2) \) is the denominator, hence the missing expression is \( 3 \).

Step 4: Understanding Rational Expression Identity

For (c), we see \( \frac{m+1}{m-3} \cdot \frac{m-5}{m-5} \). The identity property implies a multiplication by 1, thus it simplifies directly to \( \frac{m+1}{(m-3)(m-5)} \). Thus, the numerator is \( m+1 \).

Key Concepts

Multiplying Fractions Made Simple

When we multiply fractions, the process is straightforward. We multiply the numerators (the top numbers) together and the denominators (the bottom numbers) together. This creates a new fraction. Consider, for instance, the expression given in the problem, part (a):

• Start by multiplying the numerators: In \( \frac{4x}{5} \times \frac{2}{2} \), multiply 4x and 2 to get \( 8x \).
• Then, move to the denominators: Multiply 5 by 2 to end up with 10.

Putting it all together, the product is \( \frac{8x}{10} \). Remember, multiplication of fractions is all about multiplying straight across both the top and the bottom. This straightforward technique applies always, making it a handy tool in simplifying and building rational expressions.

Identity Property Explained

The identity property is a key concept in mathematics that states any number multiplied by 1 remains unchanged. This concept is pivotal when working with rational expressions. When you multiply any rational expression by a form of 1, you must keep this property in mind. In problem (b), notice \( \frac{t-2}{t-2} \). Here, we are essentially multiplying by 1 because any expression over itself equals 1. So, the calculation

• \( \frac{3}{t} \times 1 = \frac{3}{t} \)
• The denominator transforms into \( t(t-2) \)

Results in the fraction \( \frac{3}{t(t-2)} \). Using the identity property makes rational expressions easier to handle and helps preserve the original value while simplifying.

Simplifying Rational Expressions Fundamentals

Simplifying rational expressions involves reducing the expression to its simplest form while maintaining its value. To do this, we often rely on dividing both the numerator and the denominator by their common factors. This technique can be observed in part (c) of the problem:

• Given \( \frac{m+1}{m-3} \times \frac{m-5}{m-5} \), multiply the fractions using the identity property first.
• \( \frac{m-5}{m-5} = 1 \), so it doesn't change \( \frac{m+1}{m-3} \).
• Hence, simplifying leads to \( \frac{m+1}{(m-3)(m-5)} \)

This example illustrates how rational expressions can often be simplified by identifying and canceling out common factors. Mastering this skill ensures you maintain efficiency and clarity in your calculations.