Problem 48 - Algebra for College Students

Question

Solve each inequality. $$\frac{x}{x-1}>2$$

Step-by-Step Solution

Answer

The solution is \(1 < x < 2\).

Step 1: Rewrite the Inequality

Start by rewriting the given inequality. The inequality \( \frac{x}{x-1} > 2 \) can be rewritten by subtracting 2 from both sides to get a common denominator: \( \frac{x}{x-1} - \frac{2(x-1)}{x-1} > 0 \). This simplifies to \( \frac{x - 2x + 2}{x-1} > 0 \), or equivalently \( \frac{-x + 2}{x-1} > 0 \).

Step 2: Identify Critical Points

The critical points of the inequality are the points where the numerator and the denominator are zero. First, solving \(-x + 2 = 0 \) gives \(x = 2 \) and solving \(x - 1 = 0\) gives \(x = 1\). These points are critical because they could potentially change the sign of the inequality.

Step 3: Choose Test Intervals

Using the critical points 1 and 2, divide the number line into intervals: \(( -\infty, 1)\), \((1, 2)\), and \((2, \infty)\). These intervals will be used to test the sign of the expression \(\frac{-x + 2}{x-1}\) in each interval.

Step 4: Test the Intervals

1. Choose a test point from the interval \(( -\infty, 1)\), say \(x = 0\). Substituting \(x = 0\) gives \(\frac{-0 + 2}{0-1} = -2 \), which is less than 0.2. Choose a test point from the interval \((1, 2)\), say \(x = 1.5\). Substituting \(x = 1.5\) gives \(\frac{-(1.5) + 2}{1.5 - 1} = \frac{0.5}{0.5} = 1\), which is greater than 0.3. Choose a test point from the interval \((2, \infty)\), say \(x = 3\). Substituting \(x = 3\) gives \(\frac{-3 + 2}{3 - 1} = -\frac{1}{2}\), which is less than 0.

Step 5: Write the Solution

From the test points, the expression is positive in the interval \((1, 2)\). Therefore, the solution to the inequality \(\frac{x}{x-1} > 2\) is \(1 < x < 2\). Remember, we exclude the endpoints since they make the expression either zero or undefined.

Key Concepts

Critical Points in Inequalities

Critical points are the values of the variable that make either the numerator or the denominator of a rational expression equal to zero. Identifying these points is crucial because they are potential locations where the inequality's sign might change. To find the critical points, set the numerator and denominator of your expression separately to zero and solve for the variable.

For instance, given the inequality \[\frac{-x + 2}{x-1} > 0,\]we locate the critical points by solving:

• \(-x + 2 = 0\), which gives \(x = 2\).
• \(x - 1 = 0\), which results in \(x = 1\).

These points \(x = 1\) and \(x = 2\) divide the number line and help us determine where the expression changes sign. Keep in mind that the points where the denominator equals zero, such as \(x = 1\), are not included in the solution because they render the expression undefined.

Test Intervals

Once the critical points are identified, the next step is to create test intervals. These intervals segment the number line into ranges where the sign of the inequality can be consistent. This is done by placing these critical points on a number line and testing within the intervals they create.

For the inequality \[\frac{-x + 2}{x-1} > 0,\]our critical points \(x = 1\) and \(x = 2\) help create the following intervals:

• \((-\infty, 1)\)
• \((1, 2)\)
• \((2, \infty)\)

By evaluating test points within each interval, we can determine how the expression behaves in each section. This method provides insight into which intervals satisfy the inequality.

Rational Inequalities

Rational inequalities, such as \[\frac{x}{x-1} > 2,\]involve expressions where a variable appears in the numerator and the denominator. These inequalities often require more steps to solve compared to linear inequalities because they contain variable expressions that can create undefined spots or critical points.

Solving them typically involves:

• Bringing everything to one side of the inequality to compare it to zero.
• Combining the expressions over a common denominator to simplify.
• Identifying critical points which help in setting up test intervals.

This approach ensures that you capture all the complexities that may arise due to changes in the sign resulting from the variable in the denominator. It is important to always handle rational inequalities carefully to avoid possible errors due to division by zero.

Sign Analysis in Inequalities

Sign analysis plays a key role in solving inequalities, especially rational ones. It involves determining whether the rational expression is positive or negative within each test interval. This step is crucial in figuring out where the solution to the inequality lies.

For our inequality \[\frac{-x + 2}{x-1} > 0,\]we assess the sign of the entire expression within each identified interval:

• In \((-\infty, 1)\), a test point \(x = 0\) leads to a negative result \((-2)\).
• In \((1, 2)\), a test point \(x = 1.5\) results in a positive value \((1)\).
• In \((2, \infty)\), a test point \(x = 3\) gives a negative \((-\frac{1}{2})\).

Through sign analysis, we note that the inequality is satisfied in the interval where the sign is positive, which is \((1, 2)\) for this problem. This method ensures that we account for possible sign changes at critical points or where the expression is undefined.