Problem 2 - Algebra 2 and Trigonometry

Question

Gabriel wrote \(\frac{12 x}{5 x+10} \div \frac{4}{5}=\frac{12 x \div 4}{(5 x+10) \div 5}=\frac{3 x}{x+2} .\) Is Gabriel's solution correct? Justify your answer.

Step-by-Step Solution

Answer

Gabriel's solution is correct; he simplifies the fractions properly.

Step 1: Review the Original Problem

The initial expression given by Gabriel is \( \frac{12x}{5x+10} \div \frac{4}{5} \). Our task is to determine if Gabriel's solution process is correct.

Step 2: Simplify the Division of Fractions

To divide by a fraction, we multiply by its reciprocal. The reciprocal of \( \frac{4}{5} \) is \( \frac{5}{4} \). Therefore, the expression \( \frac{12x}{5x+10} \div \frac{4}{5} \) becomes \( \frac{12x}{5x+10} \times \frac{5}{4} \).

Step 3: Perform the Multiplication

Multiply the numerators and the denominators: \( \frac{12x \times 5}{(5x+10) \times 4} = \frac{60x}{20x+40} \).

Step 4: Simplify the Fraction

Factor both the numerator and the denominator. \( 60x = 2 \times 30x \) and \( 20x+40 = 20(x+2) \). Thus, the fraction simplifies to \( \frac{3x}{x+2} \) by dividing both the numerator and the denominator by 20.

Step 5: Compare to Gabriel's Solution

Gabriel's final expression is \( \frac{3x}{x+2} \), which matches the result we obtained through our calculations.

Key Concepts

Reciprocal

Understanding the concept of a reciprocal is essential when dividing fractions. The reciprocal of a fraction is simply the fraction flipped upside down.
For instance, if you have the fraction \( \frac{4}{5} \), the reciprocal is \( \frac{5}{4} \).
This is because you exchange the numerator and the denominator's positions.
When dividing fractions, you multiply by the reciprocal instead.

• This means if you are asked to divide by \( \frac{4}{5} \), you would indeed multiply by \( \frac{5}{4} \).

This method transforms tricky division operations into simpler multiplication ones, making it a handy tool in solving problems.

Simplifying Fractions

The process of simplifying fractions involves reducing fractions to their simplest form. This is done by dividing both the numerator and the denominator by their greatest common factor (GCF).
For example, if a fraction is \( \frac{60x}{20x+40} \), you first factor each term.

• The numerator is \(60x\) which can be factored into \(2 \times 30x\).
• The denominator is \(20x + 40\), which becomes \(20(x+2)\) after factoring.

Once you identify common factors, divide them out to get the simplest form.
In our scenario, both parts could be reduced by 20, resulting in \( \frac{3x}{x+2} \).
This step simplifies the fraction, making it easier to work with and interpret. It is an important skill in not just fractions but in many areas of algebra.

Multiplying Fractions

Multiplying fractions involves a straightforward process. You multiply the numerators together and then the denominators together.
Let's say your problem is \( \frac{12x}{5x+10} \times \frac{5}{4} \).

• Start by multiplying the numerators: \(12x \times 5 = 60x\).
• And then the denominators: \((5x+10) \times 4 = 20x + 40\).

The new fraction formed is \( \frac{60x}{20x+40} \).
Multiplying fractions can seem complicated, especially with variables involved, but focusing on multiplying the top numbers to get the new top, and the bottoms for the new bottom number makes it simple.
It's all about pairing the operations correctly.

Factoring Expressions

Factoring expressions is like finding a common thread within numbers that hold everything together neatly.
This process involves identifying common factors within terms to simplify or rewrite expressions.
For instance, consider the expression \(20x + 40\).

• Notice that both terms \(20x\) and \(40\) share the factor \(20\).
• By factoring this out, the expression reduces to \(20(x + 2)\).

This technique is crucial for simplifying and solving equations.
Factoring not only makes fractions easier to simplify, but it also aids in solving more complex equations by simplifying each component involved.
Once you're familiar with common factoring approaches, many algebra problems become much more manageable.