Problem 20 - College Algebra with Corequisite Support

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 \(x = \frac{4}{3}\), with \(x = -1\) and \(x = 3\) excluded.

Step 1: Identify the common denominator

The denominators in the equation are \(x+1\), \(x-3\), and \(x^2-2x-3\). We recognize \(x^2-2x-3\) can be factored into \((x-3)(x+1)\). Thus, the least common denominator (LCD) is \((x+1)(x-3)\).

Step 2: Rewrite each fraction with the common denominator

Multiply each term by \(\frac{(x+1)(x-3)}{(x+1)(x-3)}\) to express them over the common denominator:\[\frac{5}{x+1} = \frac{5(x-3)}{(x+1)(x-3)}, \quad \frac{1}{x-3} = \frac{1(x+1)}{(x+1)(x-3)}, \quad \frac{-6}{x^2-2x-3} = \frac{-6}{(x+1)(x-3)}\]

Step 3: Combine the numerators

Combine all the numerators over the common denominator:\[\frac{5(x-3) + 1(x+1) = -6}{(x+1)(x-3)}\]Simplifying the numerator gives:\[5x - 15 + x + 1 = 6x - 14\]

Step 4: Set up the equation

Since the denominators are the same, we equate the numerators:\[6x - 14 = -6\]

Step 5: Solve for \(x\)

Add 14 to both sides:\[6x - 14 + 14 = -6 + 14 \6x = 8\]Divide both sides by 6:\[x = \frac{8}{6} = \frac{4}{3}\]

Step 6: Identify excluded values

The solution must exclude the values where the original denominators are zero. For \((x+1) = 0\), \(x = -1\). For \((x-3) = 0\), \(x = 3\). Thus, \(x = -1\) and \(x = 3\) are excluded.

Key Concepts

Excluded Values

When solving rational equations, identifying excluded values at the start is crucial. Excluded values are specific numbers for the variable that would make any denominator in the equation equal to zero.
This would result in division by zero, which is undefined in mathematics.
In the original exercise, we have three denominators: \(x+1\), \(x-3\), and \(x^2-2x-3\). Here's how to find the excluded values:

• Set each distinct denominator equal to zero.
• Solve for \(x\).

For \(x+1=0\), solving gives \(x=-1\). For \(x-3=0\), solving gives \(x=3\).
Therefore, our excluded values are \(x = -1\) and \(x = 3\). These are the values that \(x\) cannot take, as they would make the denominator zero.

Common Denominator

A common denominator is often required to simplify rational equations into a single equation without fractions. The common denominator allows us to combine fractions by rewriting each term in the equation over the same denominator.
In this exercise, the denominators \(x+1\), \(x-3\), and \(x^2-2x-3\) need to be unified.Since \(x^2-2x-3\) can be factored into \((x+1)(x-3)\), the least common denominator (LCD) for all terms becomes \((x+1)(x-3)\).
With the LCD, each fraction is adjusted to reflect this common denominator:

• The fraction \(\frac{5}{x+1}\) becomes \(\frac{5(x-3)}{(x+1)(x-3)}\).
• The fraction \(\frac{1}{x-3}\) becomes \(\frac{1(x+1)}{(x+1)(x-3)}\).

Thus, the expressions can be easily combined, making the equation simpler to solve.

Factoring

Factoring is a method of rewriting a mathematical expression as a product of its simpler components. It's a fundamental technique used in solving rational equations to simplify expressions and find the common denominator.
In the given equation, \(x^2-2x-3\) is factored into two linear binomials.To factor \(x^2-2x-3\), look for two numbers that multiply to \(-3\) (the constant term) and add to \(-2\) (the coefficient of the linear term \(x\)). These two numbers are \(-3\) and \(+1\).
Hence, the expression \(x^2 - 2x - 3\) breaks down into \((x-3)(x+1)\), which matches our denominators used in forming the common denominator.Factoring helps us consolidate fractions under one denominator, aiding in the solve step of the equation effectively.

Solving Equations

Once the rational equation is organized under a common denominator, solving for \(x\) becomes straightforward because the equation now looks linear and clear. With the common denominator in place, focus shifts to the numerators.
In this problem, the fractions combine over \((x+1)(x-3)\), allowing us to focus on the numerator equation:\[6x - 14 = -6\]Next steps are algebraic simplifications:

• Add 14 to both sides to isolate the variable term: \(6x - 14 + 14 = -6 + 14\).
• Which simplifies to \(6x = 8\).
• Then, divide by 6 to solve: \(x = \frac{8}{6} = \frac{4}{3}\).

Each step systematically narrows down what \(x\) must be, as long as the solution is not one of the excluded values we identified earlier.