Problem 31 - Algebra for College Students

Question

Solve each equation. $$\frac{5}{n-3}-\frac{3}{n+3}=1$$

Step-by-Step Solution

Answer

The solutions are \(n = 6\) and \(n = -5\).

Step 1: Find a Common Denominator

Identify the common denominator for the fractions on the left-hand side of the equation. The denominators are \(n-3\) and \(n+3\). The common denominator is \((n-3)(n+3)\).

Step 2: Rewrite Each Fraction

Rewrite each fraction with the common denominator \((n-3)(n+3)\). This gives us: \(\frac{5(n+3)}{(n-3)(n+3)} - \frac{3(n-3)}{(n-3)(n+3)} = 1\).

Step 3: Combine the Fractions

Combine the fractions over the common denominator: \(\frac{5(n+3) - 3(n-3)}{(n-3)(n+3)} = 1\).

Step 4: Simplify the Numerator

Simplify the numerator: \(5(n+3) - 3(n-3) = 5n + 15 - 3n + 9 = 2n + 24\). The expression becomes: \(\frac{2n + 24}{(n-3)(n+3)} = 1\).

Step 5: Clear the Fraction

Multiply both sides of the equation by the common denominator \((n-3)(n+3)\) to clear the fractions: \(2n + 24 = (n-3)(n+3)\).

Step 6: Expand the Right-Hand Side

Expand \((n-3)(n+3)\) using the difference of squares formula: \(n^2 - 9\). So, the equation becomes: \(2n + 24 = n^2 - 9\).

Step 7: Bring All Terms to One Side

Rearrange the equation to form a quadratic equation: \(0 = n^2 - 2n - 33\). This is equivalent to \(n^2 - 2n - 33 = 0\).

Step 8: Factor the Quadratic Equation

Factor the quadratic equation: \((n-6)(n+5) = 0\).

Step 9: Solve for n

Solve for \(n\) by setting each factor equal to zero:\(n-6 = 0\), which gives \(n = 6\), and \(n+5 = 0\), which gives \(n = -5\).

Step 10: Check for Extraneous Solutions

Ensure the solutions do not conflict with the original denominators (\(n\) cannot be 3 or -3). Neither \(n=6\) nor \(n=-5\) are restricted, so both are valid solutions.

Key Concepts

Common Denominator

When dealing with equations involving fractions, finding a common denominator is an essential step. This simplifies the equation, allowing you to easily combine fractions. In our exercise, we have two fractions with denominators \(n - 3\) and \(n + 3\). To combine these fractions, we seek a common denominator, which should uniformly accommodate both expressions.

To find the common denominator, multiply the distinct factors of each denominator: \((n-3)(n+3)\). Combining these factors gives each fraction the same base, allowing them to be added or subtracted directly. Revealing a common denominator makes it possible to manipulate the equation farther along the solving process. Remember, the step of determining the common denominator is crucial as it lays the groundwork for unifying separate fractional terms. Once unified, we can focus on simplifying the numerator.

Factoring Quadratic Equations

Factoring is a key algebraic technique used to solve quadratic equations. It involves expressing a quadratic expression as the product of simpler linear factors. In our case study, after simplifying the equation, we reach a quadratic form: \(n^2 - 2n - 33 = 0\). The task here is to break down this quadratic equation into factors of the form \((n-a)(n-b)\).

To factor effectively, we seek two numbers that multiply to the constant term (\(-33\)) and add to the middle coefficient (\(-2\)). For this equation, the numbers \(-6\) and \(+5\) fit perfectly as they satisfy both conditions since \(-6 + 5 = -1\) and \(-6 \times 5 = -33\). Thus, the equation can be factored into \((n-6)(n+5) = 0\).

Factoring simplifies solving as it converts the quadratic equation into simpler linear equations. Solving the linear factors individually gives the values of \(n\) that satisfy the original quadratic equation. This makes factoring a fundamental and useful tool in algebra.

Difference of Squares

The difference of squares formula is a special algebraic identity used to expand or simplify expressions that are in the form of two square terms subtracted. The formula is \(a^2 - b^2 = (a-b)(a+b)\). This identity is quite useful for our exercise when dealing with expressions like \((n-3)(n+3)\).

In our rational equation, using the difference of squares for the denominator simplifies the product \((n-3)(n+3)\) to \(n^2 - 9\). Recognizing and applying the difference of squares efficiently reduces the complexity of the equation, allowing us to solve it more directly.

Identifying difference of squares helps us to expand or factor equations quickly which expedites solving and highlights any simplifications present. This technique is commonly employed in algebra to handle binomials efficiently.