Problem 1 - Algebra 2 and Trigonometry

Question

Samantha said that the equation \(\frac{a-2}{a}=\frac{a+2}{2 a}\) in Example 2 could be solved by multiplying both sides of the equation by 2\(a\) . Would Samantha's solution be the same as the solution obtained in Example 2\(?\) Explain why or why not.

Step-by-Step Solution

Answer

Yes, Samantha's method would yield the same solution of \(a = 6\). It effectively removes the fractions, simplifying the process.

Step 1: Understand the Equation

The given equation is \( \frac{a-2}{a} = \frac{a+2}{2a} \). Both terms are rational expressions and have denominators of \(a\) and \(2a\) respectively, resulting in the inequality.

Step 2: Multiply by Least Common Multiple

Samantha suggests multiplying both sides by \(2a\), which is the least common multiple of the denominators (\(a\) and \(2a\)). This removes the fractions:\[ 2a \times \frac{a-2}{a} = 2a \times \frac{a+2}{2a} \].

Step 3: Simplify Both Sides

On the left side, \(2a\) and \(a\) cancel out to leave \(2(a - 2)\). On the right side, \(2a\) and \(2a\) completely cancel out to leave \(a + 2\). Thus, simplifying results in:\[ 2(a - 2) = a + 2 \].

Step 4: Solve the Simpler Equation

Distribute and simplify:\[ 2a - 4 = a + 2 \].Subtract \(a\) from both sides:\[ a - 4 = 2 \].Add 4 to both sides:\[ a = 6 \].

Step 5: Compare to Example 2

In Example 2, solving the original equation using other methods should give \(a = 6\) as well. Multiplying by \(2a\) is simply a different method to reach the same solution efficiently.

Key Concepts

Least Common Multiple

In the world of rational equations, it's common to encounter fractions. Fractions can make equations seem more complex than they actually are. This is where the concept of the Least Common Multiple (LCM) becomes an essential tool. The LCM of two numbers is the smallest number that both numbers can divide into without leaving a remainder.

When dealing with rational equations like \( \frac{a-2}{a} = \frac{a+2}{2a} \), identifying the LCM of the denominators helps to eliminate the fractions. Here, the denominators are \(a\) and \(2a\). The LCM of \(a\) and \(2a\) is \(2a\).

By multiplying both sides of the equation by \(2a\), each term's denominator is effectively removed, simplifying the equation into a more manageable form. For a rational equation, this technique simplifies the expression, turning a seemingly complex equation into a simpler linear form where standard solving techniques apply.

• This process helps to avoid errors usually caused by dealing with multiple fractions.
• It provides a straightforward way to simplify the problem early on.

Understanding how to determine and use the LCM is crucial in rational equations. It builds a foundation that allows students to approach complex problems with confidence.

Cross Multiplication

Cross multiplication is another technique used in solving rational equations. It's quite handy when dealing with an equation set up like two fractions equaling each other. This method allows us to steer away from the fractions by simply multiplying across the equality.

In the exercise, taking \( \frac{a-2}{a} = \frac{a+2}{2a} \) as an example, cross multiplication involves multiplying the numerator of one fraction by the denominator of the other, and setting those products equal to each other:

• Multiply \( (a - 2) \) by \( 2a \)
• Multiply \( (a + 2) \) by \( a \)

Cross multiplication gives us: \[ (a - 2) \cdot 2a = (a + 2) \cdot a \]This process bypasses the division in rational expressions, converting the equation into a linear form. This is particularly useful because linear equations are easier to manage and solve. Cross multiplication offers a clear road to simplifying equations and avoiding the common pitfalls of fraction arithmetic. Students find this method intuitive once practiced, as it’s like setting a balance between the two sides by crossing them over.

Solving Linear Equations

Once rational expressions have been simplified, they usually transform into linear equations. Solving linear equations is a fundamental skill in algebra that requires understanding how to manipulate the equation to isolate the variable.

After using techniques like LCM or cross multiplication, the equation \( 2(a - 2) = a + 2 \) was obtained. The goal here is to solve this equation by isolating \( a \).

Here's how you can solve it step-by-step:

• First, distribute to simplify both sides. This means multiplying through any parentheses: \( 2a - 4 = a + 2 \).
• Next, get all terms containing \( a \) on one side of the equation. You can do this by subtracting \( a \) from both sides, resulting in: \( a - 4 = 2 \).


• The final step is to isolate \( a \) completely. Add 4 to both sides to find that \( a = 6 \).

What’s significant here is showing how applying these methods turns a perceived complex issue into a simple arithmetic manipulation. Understanding how to simplify equations helps students tackle more advanced mathematics with ease and confidence.