Problem 20 - Intermediate Algebra

Question

Solve equation. \(\frac{1}{m+3}-\frac{m}{m^{2}-9}=\frac{-2}{m-3}\)

Step-by-Step Solution

Answer

No solution; \(m = -3\) is not valid because it makes the equation undefined.

Step 1: Understand the Problem

The given equation \( \frac{1}{m+3}-\frac{m}{m^2-9}=\frac{-2}{m-3} \) involves fractions with denominators \( m+3 \), \( m^2-9 \), and \( m-3 \). The denominator \( m^2-9 \) factors as \((m+3)(m-3)\). So, the common denominator for all terms is \((m+3)(m-3)\).

Step 2: Rewrite Fractions with Common Denominator

Rewrite each fraction with the common denominator \((m+3)(m-3)\). The first term becomes \( \frac{1 \cdot (m-3)}{(m+3)(m-3)} \), the second term is already \( \frac{m}{(m^2-9)} \), and the third term becomes \( \frac{-2 \cdot (m+3)}{(m+3)(m-3)} \).

Step 3: Simplify the Numerator

Combine the numerators over the common denominator: \( \frac{m-3 - m - 2(m+3)}{(m+3)(m-3)} \). Expand and simplify the numerator: \( m - 3 - m - 2m - 6 \). This simplifies to \(-3m - 9\).

Step 4: Set the Numerator Equal to Zero

For the fraction to equal zero, the numerator must be zero. So, solve \(-3m - 9 = 0\).

Step 5: Solve the Equation

Add 9 to both sides to get \(-3m = 9\). Then divide both sides by \(-3\) to find \(m = -3\).

Step 6: Check for Extraneous Solutions

Check if \(m = -3\) is valid by substituting back into the denominators. Since \(m+3\) becomes zero, this makes the original equation undefined at \(m = -3\). So, it is not a valid solution.

Key Concepts

Understanding the Common Denominator

When dealing with rational equations, a key step is finding a common denominator. This allows you to combine terms and simplify the equation more effectively. In the equation \( \frac{1}{m+3}-\frac{m}{m^2-9}=\frac{-2}{m-3} \), the denominators are \( m+3 \), \( m^2-9 \), and \( m-3 \). To find the common denominator:

• Notice that \( m^2-9 \) can be factored into \((m+3)(m-3)\).
• This makes \((m+3)(m-3)\) the least common denominator, as it is a multiple of all the others.

By rewriting each fraction with this common denominator, you facilitate the process of combining and simplifying the rational expression. This step is crucial before moving on to appropriately manipulations, such as simplifying or solving the equation.

Factoring Polynomials for Simplification

Factoring is a powerful tool in algebra and is particularly helpful when working with polynomials. It involves breaking down a polynomial into simpler pieces or expressions, which are multiplied together to result in the original polynomial.For the given rational equation, notice the denominator \( m^2-9 \), which is a difference of squares. It can be factored as follows:

• Recognize \( m^2-9 \) as \((m+3)(m-3)\), because \( a^2-b^2 = (a-b)(a+b) \).

Factoring helps identify the common denominator and assists in simplifying expressions. By converting \( m^2-9 \) to its factored form, we seamlessly set up a workable equation to combine the fractions, paving the way for further simplification.

Recognizing and Handling Extraneous Solutions

Extraneous solutions can occur in algebra when you perform operations that have the potential a solution might satisfy a transformed equation but not the original one. It's particularly important in rational equations where variables in the denominator can result in undefined expressions.After solving the simplified equation \(-3m - 9 = 0\), we find that \( m = -3 \). However:

• Substituting \( m = -3 \) back into the original denominators \( m+3 \) results in zero, making the fraction undefined.

Due to this, \( m = -3 \) cannot be a solution to the equation. The term extraneous refers to solutions arising from the solving process that don't satisfy the original equation. It's essential to verify solutions in the context of the original problem, particularly when dealing with rational equations to avoid such pitfalls.