Problem 10 - Algebra for College Students

Question

Solve each equation. $$ 3+\frac{9}{n-3}=\frac{27}{n^{2}-3 n} $$

Step-by-Step Solution

Answer

\(n = -3\) is the solution.

Step 1: Establish Common Denominator

The equation is \(3 + \frac{9}{n-3} = \frac{27}{n^{2} - 3n}\). The least common denominator between the fractions is \(n(n-3)\). We rewrite \(3\) as \(\frac{3n(n-3)}{n(n-3)}\) so that it can be combined with the other fractions.

Step 2: Clear the Fractions

Multiply every term in the equation by the common denominator \(n(n-3)\) to eliminate the fractions:\[ 3n(n-3) + 9n = 27 \].Simplify to get a polynomial equation.

Step 3: Simplify and Rearrange

Expand the terms and simplify:\[ 3n^2 - 9n + 9n = 27 \].This simplifies to:\[ 3n^2 = 27 \].

Step 4: Solve for \(n\)

Divide each side by 3 to isolate the squared term:\[ n^2 = 9 \].Take the square root of both sides of the equation:\[ n = 3 \] or \[ n = -3 \].

Step 5: Check for Extraneous Solutions

Substitute \(n = 3\) back into the denominators \(n-3\) and \(n^2 - 3n\). The denominators become zero, which means \(n = 3\) is not a valid solution. Check \(n = -3\): Substitute it back to ensure it doesn't make denominators zero:- \(n = -3\) makes the denominators non-zero, so it is valid.

Key Concepts

The Importance of Finding a Common Denominator

A common denominator is crucial when solving equations involving fractions, especially when the fractions have different denominators. In our given problem, we need to deal with the fractions \(\frac{9}{n-3}\) and \(\frac{27}{n^2-3n}\). The key is to find a common denominator that will allow us to combine these fractions.

• First, identify the denominators: \(n-3\) and \(n^2-3n\).
• Factor the second denominator: \(n^2-3n = n(n-3)\).
• The least common denominator (LCD) is \(n(n-3)\), the product of all unique factors.

Rewriting each term over the LCD ensures that all the fractions have the same bottom part. Then, multiplying through by this common denominator lets us clear the fractions entirely, simplifying our work.

Understanding Polynomial Equations

A polynomial equation consists of variables raised to whole-number powers, coefficients, and constants. In our problem, after clearing the fractions, we arrive at the equation \(3n^2 - 9n + 9n = 27\), which simplifies to \(3n^2 = 27\).

• This is a simple quadratic equation: \(n^2 = 9\) after dividing by 3.
• Quadratics typically take the form \(ax^2 + bx + c = 0\) and can be solved by factoring, completing the square, or using the quadratic formula.

Quadratic equations often yield two solutions, as seen when taking the square root of both sides in \(n^2 = 9\), resulting in \(n = 3\) or \(n = -3\). Solving polynomial equations requires recognizing the highest degree term and systematically breaking down the equation.

Recognizing and Avoiding Extraneous Solutions

Extraneous solutions are results from solving an equation that are invalid or do not satisfy the original equation. They often appear during algebraic manipulations, especially when both sides are squared or when clearing fractions. In this exercise, we found solutions \(n = 3\) and \(n = -3\).

• Check each potential solution by substituting back into the original denominators.
• If a solution causes any denominator to become zero, it's extraneous.

For \(n = 3\), substituting back into \(n-3\) yields 0, creating an undefined fraction, hence \(n = 3\) is extraneous. Verifying \(n = -3\) results in valid expressions, confirming it as a legitimate solution. Always ensure solutions don't break the fundamental rules of algebra that apply to the problem at hand.