Problem 49 - College Algebra Essentials

Question

Contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind , solve the equation. \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2 x-8}\)

Step-by-Step Solution

Answer

The solution to the equation \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{x^{2}-2 x-8}\) is \(x=14\). The restrictions on the variable are \(x \neq 4\) and \(x \neq -2\).

Step 1: Identify the Restrictions

Set each of the denominators equal to zero and solve for x. This will give the restrictions on the variable x. \[x-4=0\] gives \(x=4\) and \[x+2=0\] gives \(x=-2\]. These are the restrictions, so \(x\) cannot be 4 or -2.

Step 2: Combine Like Terms

To solve this equation let's first simplify the right hand side. We observe that the denominator \(x^{2}-2 x-8\) is nothing but \((x-4)(x+2)\), which are the denominators on the left hand side. So, rewrite the equation as: \(\frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{(x-4)(x+2)}\), and simplify by multiplying both sides of the equation by \((x-4)(x+2)\) which is simply \(x^{2}-2 x-8\). Doing this, we get: \(x+2 - 5(x-4) = 6\), simplify the equation to: \(x-20=-6\).

Step 3: Solve for x

Finally, to determine the value of x, add 20 to both sides to isolate x, which gives: \(x = -6 + 20 = 14\)

Step 4: Check the Solution

Remember the restrictions from Step 1. Our solution is \(x=14\), which does not violate these restrictions. Therefore, \(x=14\) is the solution to the given equation

Key Concepts

Understanding Denominators in Rational Equations

In rational equations, denominators play a crucial role. They are the bottom part of a fraction and dictate the division of the numerator by the value below. The essence of denominators:

• They affect the existence of the solutions.
• If a denominator is zero, the fraction becomes undefined.

In our equation \( \frac{1}{x-4} - \frac{5}{x+2} = \frac{6}{x^2-2x-8} \), there are denominators \( x-4 \), \( x+2 \), and \( x^2-2x-8 \). To manage these, we factor the quadratic denominator when possible to identify shared components. This helps later during simplification. Correct handling of denominators is essential for accurate solving.

Identifying Variable Restrictions

Variable restrictions arise when certain values make a denominator zero, as division by zero is undefined in mathematics.Finding restrictions:

• Set each individual denominator to zero.
• Solve for the variable to find restrictions.

For example, setting \( x-4 = 0 \) results in \( x = 4 \). Similarly, \( x+2 = 0 \) leads to \( x = -2 \). These values are our restrictions, meaning \( x \) cannot equal these numbers. Ignoring these would lead to incorrect solutions, so always factor them into your solution strategy.

Techniques for Solving Equations

The goal of solving rational equations is to find what value of the variable makes the equation true, while adhering to any restrictions. Steps to solve:

• Simplify the equation when possible, for instance by factoring.
• Get rid of the denominators by multiplying through by the least common denominator (LCD).

In our problem, we used the factored form \( (x-4)(x+2) \) to eliminate fractions.After simplification, solve the resulting linear equation. In this case, it reduced to \( x-20 = -6 \), and solving gives \( x = 14 \).This approach ensures solutions meet algebraic expectations.

Factoring Quadratics for Simplification

Factoring is key in rearranging expressions to a simpler form, making it particularly handy in rational equations.Why factor?

• Makes complex expressions manageable.
• Reveals values that make the equation undefined (restrictions).

Quadratics like \( x^2-2x-8 \) factor to \((x-4)(x+2)\). Identifying these factors not only helps consolidate terms but assists in finding variable restrictions and in the simplification process, as common factors can be cancelled to simplify the equation further.Mastering this skill improves problem-solving efficiency significantly.