Problem 18 - 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. $$ \frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{(x-1)(x-2)} $$

Step-by-Step Solution

Answer

The solution is \(x = 4\), with exclusions \(x = 1\) and \(x = 2\).

Step 1: Identify Denominators and Exclusions

To solve the equation \(\frac{3}{x-2} = \frac{1}{x-1} + \frac{7}{(x-1)(x-2)}\), first identify the denominators: \(x-2\), \(x-1\), and \((x-1)(x-2)\). Values of \(x\) that make any denominator zero are excluded. Setting each denominator to zero, find that \(x = 2\) and \(x = 1\) are the excluded values because they cause division by zero.

Step 2: Find a Common Denominator

Find the least common denominator (LCD) for the terms, which is \((x-1)(x-2)\). This combines all factors in the denominators.

Step 3: Clear the Fractions

Multiply every term by the LCD, \((x-1)(x-2)\), to clear the fractions from the equation: \[(x-1)(x-2) \times \frac{3}{x-2} = (x-1)(x-2) \times \frac{1}{x-1} + (x-1)(x-2) \times \frac{7}{(x-1)(x-2)}\] Simplifying, this results in: \[3(x-1) = (x-2) + 7\] which simplifies further to an equation without fractions.

Step 4: Simplify and Solve the Equation

Simplify the resulting equation: \[3x - 3 = x - 2 + 7\]Combine like terms to get: \[3x - 3 = x + 5\]Subtract \(x\) from both sides: \[2x - 3 = 5\]Add 3 to both sides: \[2x = 8\]Divide both sides by 2 to solve for \(x\): \[x = 4\].

Step 5: Verify and State the Solution Set

Verify that \(x=4\) does not make any denominator zero (i.e., not in the excluded values \(x=1\) or \(x=2\)). Therefore, 4 is the solution, and it is valid.

Key Concepts

Excluded Values

In solving rational equations, it is crucial to identify the excluded values. These are specific values of the variable that can make the denominator zero, resulting in an undefined expression. Anytime you have a fraction, division by zero is undefined, hence the need to exclude these values from the solution.

In the provided equation \(\frac{3}{x-2} = \frac{1}{x-1} + \frac{7}{(x-1)(x-2)}\), let's consider the denominators closely for any zero values:

• \(x-2 = 0\) implies \(x = 2\).
• \(x-1 = 0\) implies \(x = 1\).
• Any common denominator like \((x-1)(x-2)\) carries these same exclusions.

Therefore, both \(x = 1\) and \(x = 2\) must be excluded from the solution set because they would make the original denominators zero, leading to undefined operations.

Common Denominator

Finding a common denominator is essential for solving rational equations because it allows you to eliminate the fractions and simplify the equation. To combine fractions into a single expression or to equate them, they must have a like denominator.

In the example equation \(\frac{3}{x-2} = \frac{1}{x-1} + \frac{7}{(x-1)(x-2)}\), the denominators are \(x-2\), \(x-1\), and \((x-1)(x-2)\).

• TheLCD (Least Common Denominator) is \((x-1)(x-2)\), as it incorporates all factors from the different denominators.

By expressing each term with this common denominator, you can then clear the fractions, making the subsequent algebraic manipulation much more straightforward.

Equation Simplification

Once you have a common denominator, the goal is to simplify the equation by removing the fractions. Use the common denominator to multiply across all terms, ensuring that you are clear of fractional expressions.

Here is how it works for our equation:

• Multiply every term by the LCD \((x-1)(x-2)\).
• The equation becomes \[3(x-1) = (x-2) + 7\].

After eliminating the fractions, simplify by distributing and collecting like terms. This process turns your equation into a simpler form without fractions, and involves straightforward algebra steps of expanding, rearranging, and combining like terms to reach a solution.

Fractions

Fractions are an integral part of rational equations, and understanding them is necessary when working with such equations. Each fraction in a rational equation represents a division of the numerator by the denominator, which must be non-zero to be defined.

Here's a helpful breakdown for our equation:

• Each fraction needs to be carefully rewritten during the common denominator step.
• Understanding the behavior and form of fractions will help in clearing them correctly by multiplying through the entire equation with the common denominator.

Once fractions are eliminated, the equation becomes far simpler, allowing you to use basic algebra skills to find the solution directly, as demonstrated in the simplification and solving steps of the exercise.