Problem 20 - College Algebra

Question

For the following exercises, solve each rational equation for \(x\). State all \(x\) -values that are excluded from the solution set. $$ \frac{5}{x+1}+\frac{1}{x-3}=\frac{-6}{x^{2}-2 x-3} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{4}{3}\); exclude \(x = -1\) and \(x = 3\).

Step 1: Factor the Denominator

First, we need to factor the denominator of the right-hand side of the equation, which is \(x^2 - 2x - 3\). Notice that this is a quadratic expression, which can be factored into \((x - 3)(x + 1)\). This gives us the equation \[\frac{5}{x+1} + \frac{1}{x-3} = \frac{-6}{(x+1)(x-3)}.\]

Step 2: Identify Excluded Values

The excluded values are found by setting each factor in the denominator equal to zero. From \(x+1=0\), we get \(x=-1\). From \(x-3=0\), we get \(x=3\). These are the values where the denominator becomes zero, so they are excluded from the solution set: \(x = -1\) and \(x = 3\).

Step 3: Clear Denominators

Multiply every term in the equation by the least common denominator, which is \((x+1)(x-3)\), to eliminate the fractions: \[(x+1)(x-3)\left(\frac{5}{x+1}\right) + (x+1)(x-3)\left(\frac{1}{x-3}\right) = (x+1)(x-3)\left(\frac{-6}{(x+1)(x-3)}\right).\] Simplifying, we have \[5(x-3) + 1(x+1) = -6.\]

Step 4: Simplify the Equation

Distribute the terms inside the parentheses: \[5x - 15 + x + 1 = -6.\] Combine like terms to get \[6x - 14 = -6.\]

Step 5: Solve for x

Add 14 to both sides of the equation: \[6x = 8.\] Now divide both sides by 6 to solve for \(x\): \[x = \frac{8}{6} = \frac{4}{3}.\]

Step 6: Verify Exclusion

Verify that \(x = \frac{4}{3}\) is not one of the excluded values \(x = -1\) or \(x = 3\). Since \(\frac{4}{3}\) does not equal either of these values, it is a valid solution to the equation.

Key Concepts

Factoring Quadratics

Factoring quadratics is an essential skill when dealing with rational equations. In this problem, we notice the quadratic expression in the denominator: \[x^2 - 2x - 3\].To factor it, we need to express it as a product of two binomials.The process involves finding two numbers that multiply to the constant term (-3 here)and add up to the linear coefficient (-2 here). These numbers are -3 and +1. Thus, the factored form is:

• \[(x - 3)(x + 1)\].

By turning the quadratic into a product of factors, we can more easily identify solutions and manage denominators, as seen in solving rational equations involving quadratics.

Excluded Values

Excluded values are crucial when working with rational equations. These are the values that make the denominator zero, leading to undefined expressions. To find the excluded values, set each factor of the denominator to zero:

• \[x + 1 = 0\] implies \[x = -1\].
• \[x - 3 = 0\] implies \[x = 3\].

These values \(x = -1\) and \(x = 3\) are excluded from the solution set because they make the denominators of terms in the equation undefined. Always check for and state excluded values before proceeding with solving rational equations to avoid erroneous solutions.

Solving Equations

Solving rational equations involves several steps. After identifying and factoring the denominators, the next step is to solve the equation by eliminating fractions. This is done by multiplying each term by the Least Common Denominator (LCD), which in this case, is \[(x + 1)(x - 3)\].This process eliminates the fractions and turns the equation into a simpler linear form. Follow these steps:

• Expand and distribute terms: \[5(x - 3) + 1(x + 1) = -6\].
• Simplify: \[5x - 15 + x + 1 = -6\].
• Combine like terms: \[6x - 14 = -6\].

Finally, solve for \(x\) by isolating it. Add 14 to both sides: \[6x = 8\].Divide both sides by 6:\[x = \frac{4}{3}\].This gives us the solution in its simplest form.

Least Common Denominator

The Least Common Denominator (LCD) is key in solving rational equations. It is the smallest expression that each of the denominators can divide into without a remainder. For the given rational equation, the LCD would be the expression that includes all factors from the denominators: \[(x + 1)(x - 3)\].The LCD simplifies the process of clearing fractions as it encompasses all necessary terms. Multiply each part of the equation by the LCD to transition from fractions to equations with whole expressions:

• This step transforms the original equation effectively, making it simpler to solve.
• It involves entirely removing the denominators, enabling straightforward algebraic manipulation.

Using the LCD effectively bridges the gap from complex fraction-based problems to solving simple equations.