Problem 13 - Elementary Algebra

Question

For the following problems, simplify the expressions. $$ \frac{8 \cdot 6}{2}+\frac{9 \cdot 9}{3}-\frac{10 \cdot 4}{5} $$

Step-by-Step Solution

Answer

Question: Simplify the expression: $$ \frac{8 \cdot 6}{2} + \frac{9 \cdot 9}{3} - \frac{10 \cdot 4}{5} $$ Answer: 43

Step 1: Simplify each fraction

First, we will simplify the numerator and denominator of each fraction: $$ \frac{8 \cdot 6}{2} = \frac{48}{2} , \frac{9 \cdot 9}{3} = \frac{81}{3} , \frac{10 \cdot 4}{5} = \frac{40}{5} $$

Step 2: Calculate simplified fractions

Now, we will calculate each fraction individually: $$ \frac{48}{2} = 24 , \frac{81}{3} = 27 , \frac{40}{5} = 8 $$

Step 3: Perform addition and subtraction operations

With the simplified fractions, we can perform the addition and subtraction operations: $$ 24 + 27 - 8 $$

Step 4: Calculate the final result

Now perform the addition and subtraction: $$ 24 + 27 - 8 = 43 $$ Thus, the simplified expression is equal to 43.

Key Concepts

Understanding Fractions

Fractions are a way to represent parts of a whole. Any fraction consists of a numerator and a denominator. The fraction itself is a form of division, indicating how the numerator is divided by the denominator.

When working with fractions in algebra, it is crucial to understand how they interact in arithmetic operations like addition and subtraction. Simplifying fractions is often the first step to make these operations easier. By simplifying fractions, you express them in their simplest form, which means the numerator and denominator have no common factors other than 1.

When looking at the original exercise, we see fractions such as \(\frac{8 \cdot 6}{2}\). By simplifying the fraction, we can more easily perform additional math operations. Using fractions efficiently helps in both simplifying calculations and solving complex algebraic expressions.

What is the Numerator?

The numerator is the top number in a fraction. It represents the number of equal parts you have. For example, in the fraction \(\frac{8 \cdot 6}{2}\), the numerator is \(8 \cdot 6 = 48\).

To simplify expressions, calculating the numerator's product first is often a key step. Ensuring that you correctly compute this can determine the accuracy of your final solution. In our exercise, each fraction's numerator is simplified by multiplication.

• For \(\frac{8 \cdot 6}{2}\), the numerator is 48.
• For \(\frac{9 \cdot 9}{3}\), the numerator is 81.
• For \(\frac{10 \cdot 4}{5}\), the numerator is 40.

By clearly understanding and accurately calculating numerators, you lay the foundation for solving the expression correctly.

Getting to Know Denominators

The denominator is the bottom number in a fraction. It indicates into how many equal parts the whole is divided. In the context of our problem, the denominators were naturally numbers looking for reduction.

Simplifying a fraction involves dividing the numerator by the denominator. The exercise example assumes we simplify \(\frac{48}{2}\), \(\frac{81}{3}\), and \(\frac{40}{5}\), yielding results of 24, 27, and 8 respectively.

• The denominator for \(\frac{8 \cdot 6}{2}\) is 2.
• The denominator for \(\frac{9 \cdot 9}{3}\) is 3.
• The denominator for \(\frac{10 \cdot 4}{5}\) is 5.

Understanding the role of denominators helps simplify fractions, which is crucial for making further complex algebraic operations manageable.

Addition and Subtraction in Algebra

Once the fractions are simplified, addition and subtraction are performed just like with whole numbers. This step involves combining the simplified results to find the final answer.

In algebra, it's important to carry out these operations in a systematic order. From the exercise, once fractions are simplified to numbers, the operation becomes simpler:

• After simplification, we calculate: \(24 + 27 - 8\).
• We do this step by step to ensure accuracy: first, add 24 and 27, getting 51.
• Then subtract 8 from 51, resulting in a final answer of 43.

These basic operations in algebra help transition complex algebraic expressions into simpler forms, making it easier for students to solve and understand the underlying mathematics.