Problem 19
Question
For the following exercises, solve each rational equation for \(x .\) State all \(x\) -values that are excluded from the solution set. $$ \frac{3 x}{x-1}+2=\frac{3}{x-1} $$
Step-by-Step Solution
Verified Answer
No solution because \( x = 1 \) is excluded.
1Step 1: Identify the Denominator Restrictions
Examine the denominators in the equation \( \frac{3x}{x-1} + 2 = \frac{3}{x-1} \). The denominator is \( x-1 \), which must not be zero. This implies that \( x-1 eq 0 \), so \( x eq 1 \). The value \( x = 1 \) is excluded from the solution set.
2Step 2: Eliminate the Denominators
To eliminate the denominators, multiply every term by \( x-1 \) to clear the fractions. The equation becomes: \[ (x-1) \left( \frac{3x}{x-1} \right) + 2(x-1) = (x-1) \left( \frac{3}{x-1} \right) \] Simplifying gives: \[ 3x + 2(x-1) = 3 \]
3Step 3: Simplify and Combine Terms
Distribute the \(2\) in the equation: \[ 3x + 2x - 2 = 3 \] Then combine the terms with \(x\): \[ 5x - 2 = 3 \]
4Step 4: Solve for x
Add \(2\) to both sides of the equation:\[ 5x = 5 \] Now divide by \(5\) to solve for \(x\):\[ x = 1 \]
5Step 5: Check Exclusion
Check if the solution \(x = 1\) is part of the exclusions. Since \(x = 1\) was excluded in Step 1 due to making the denominator zero, there are no valid solutions for the equation.
Key Concepts
Denominator RestrictionsExclusion of ValuesClearing Fractions
Denominator Restrictions
In any rational equation, it's crucial to focus on denominators. The denominators determine the restrictions for variable values. Let's consider the equation \( \frac{3x}{x-1} + 2 = \frac{3}{x-1} \). Notice that both fractions have \( x-1 \) as the denominator.
By ensuring that the denominator never equals zero, we can pinpoint values of \(x\) that must be excluded to avoid undefined expressions. Here, since \( x-1 \) potentially becomes zero when \( x = 1 \), the value \( x = 1 \) is restricted.
Therefore, it is essential to note and exclude this value from possible solutions to maintain the legitimacy of the equation.
When dealing with any rational equation, always remember to:
By ensuring that the denominator never equals zero, we can pinpoint values of \(x\) that must be excluded to avoid undefined expressions. Here, since \( x-1 \) potentially becomes zero when \( x = 1 \), the value \( x = 1 \) is restricted.
Therefore, it is essential to note and exclude this value from possible solutions to maintain the legitimacy of the equation.
When dealing with any rational equation, always remember to:
- Identify all possible denominators.
- Set each denominator not equal to zero.
- Solve these inequalities to determine restrictions.
Exclusion of Values
After identifying denominator restrictions, you have to exclude certain values from the solution set. In rational equations, values that make any denominator zero cannot be solutions. This step requires careful examination and re-evaluation of potential solutions.
In our example, the value \( x = 1 \) must be excluded because it makes the denominator zero, as you determined when applying denominator restrictions. When you solve the equation and arrive at a solution, always check this result against excluded values.
For example:
In our example, the value \( x = 1 \) must be excluded because it makes the denominator zero, as you determined when applying denominator restrictions. When you solve the equation and arrive at a solution, always check this result against excluded values.
For example:
- Solve the equation completely.
- Double-check that your solution is not an excluded value.
Clearing Fractions
Once you've identified the exclusions, the next step is clearing fractions to simplify the equation. By clearing fractions, you turn a complex equation into a simpler linear or polynomial one.
In our scenario, after recognizing and excluding \( x = 1 \), multiply each term of the equation \( \frac{3x}{x-1} + 2 = \frac{3}{x-1} \) by \( x-1 \). This process eliminates the denominators:
\[(x-1)\left(\frac{3x}{x-1}\right) + 2(x-1) = (x-1)\left(\frac{3}{x-1}\right)\]
It simplifies to:
\[3x + 2(x-1) = 3\]
By removing fractions, you convert the rational equation into an easily solvable form. Simplicity in this method streamlines the path to the final solution, leading to accurate and manageable results.
To clear fractions effectively:
In our scenario, after recognizing and excluding \( x = 1 \), multiply each term of the equation \( \frac{3x}{x-1} + 2 = \frac{3}{x-1} \) by \( x-1 \). This process eliminates the denominators:
\[(x-1)\left(\frac{3x}{x-1}\right) + 2(x-1) = (x-1)\left(\frac{3}{x-1}\right)\]
It simplifies to:
\[3x + 2(x-1) = 3\]
By removing fractions, you convert the rational equation into an easily solvable form. Simplicity in this method streamlines the path to the final solution, leading to accurate and manageable results.
To clear fractions effectively:
- Identify a common multiplier, typically the least common denominator.
- Multiply each term by this value, ensuring denominators disappear.
- Simplify and continue solving as a regular equation.
Other exercises in this chapter
Problem 19
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