Problem 40 - College Algebra

Question

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x}{x-1}>2 $$

Step-by-Step Solution

Answer

The solution to the inequality \(\frac{x}{x - 1} > 2\) in interval notation is (1, 2).

Step 1: Simplification

Subtract 2 from both sides of the inequality to simplify it: \[\frac{x}{x-1}-2>0\]. Next, in order to subtract fractions, they should have the same denominator. The denominator of 2 could be considered as (x - 1) by multiplying 2 with (x - 1) to get 2(x - 1). This results in \[\frac{x}{x - 1} - \frac{2(x - 1)}{x - 1} > 0\] which simplifies to \[\frac{x - 2x + 2}{x -1} > 0\] or \[\frac{2 - x}{x - 1} > 0.\]

Step 2: Finding Critical Points

The critical points are obtained by setting the numerator and the denominator equal to 0 and then solving for x. \[2 - x = 0 \rightarrow x = 2\] and \[x -1 = 0 \rightarrow x = 1.\] So the critical points are x = 1 and x = 2.

Step 3: Interval Test

Now the real number line is divided into three intervals by the critical points: (-∞, 1), (1, 2) and (2, ∞). Choose a test point in each interval and substitute it into the simplified inequality. If the result is true, that interval is part of the solution. For (-∞, 1), let x = 0: \[\frac{2 - 0}{0 - 1} > 0 \rightarrow -2 > 0\] which is false so (-∞, 1) is not part of the solution. For (1, 2), let x = 1.5: \[\frac{2 - 1.5}{1.5 - 1} > 0 \rightarrow 1 > 0\] which is true so (1, 2) is part of the solution. For (2, ∞), let x = 3: \[\frac{2 - 3}{3 - 1} > 0 \rightarrow -0.5 > 0\] which is false so (2, ∞) is not part of the solution.

Step 4: Express Solution in Interval Notation

The solution to the inequality \(\frac{x}{x - 1} > 2\) is the interval that satisfied the inequality which is (1, 2). Based on the inequality sign '>', it shows that it is 'greater than' but not 'equal to', so the solution does not include the endpoints. Thus, the final answer in interval notation is (1, 2).

Key Concepts

Interval Notation

Interval notation is a way of expressing a solution set for inequalities. It uses parentheses and brackets to show which numbers are included in the set.
Parentheses \((, )\) mean the endpoints are not included, while brackets \([, ]\) mean they are included.
For example, the interval \((1, 2)\) includes all numbers greater than 1 and less than 2, but not 1 or 2 themselves. This is useful because it provides a clear, concise way to describe the numbers that satisfy an inequality.

• Use open intervals for inequalities with < or >, like \(x > 1\).
• Use closed intervals for inequalities with \(\leq\) or \(\geq\), like \(x \leq 3\).

When you graph the solution on a number line, open circles are used at the boundaries for open intervals, and filled circles for closed intervals. This visual representation helps in understanding which numbers belong to the solution set.

Critical Points

Critical points in rational inequalities are values where the expression equals zero or becomes undefined. These points help divide the number line into intervals to test for solutions.
To find critical points:

• Set the numerator equal to zero and solve for \(x\). This determines where the fraction is zero.
• Set the denominator equal to zero as it indicates where the expression is undefined.

For instance, in the inequality \(\frac{2 - x}{x - 1} > 0\), set \(2 - x = 0\) which gives \(x = 2\), and \(x - 1 = 0\) which gives \(x = 1\).
These critical points \(x = 1\) and \(x = 2\) are crucial as they mark changes in the sign of the inequality. They are used to divide the number line into intervals for further analysis.

Test Intervals

Once critical points divide the number line, test intervals help determine which sections satisfy the inequality. To do this:
1. Choose a test point from each interval created by the critical points.
2. Substitute the test point back into the inequality.
3. Check if the inequality is true or false for that interval.

• For \((-\infty, 1)\), test with \(x = 0\): \(\frac{2 - 0}{0 - 1} = -2\), which is false.
• For \((1, 2)\), test with \(x = 1.5\): \(\frac{2 - 1.5}{1.5 - 1} = 1\), which is true.
• For \((2, \infty)\), test with \(x = 3\): \(\frac{2 - 3}{3 - 1} = -0.5\), which is false.

Only the interval \((1, 2)\) satisfies the inequality \(\frac{2 - x}{x - 1} > 0\). This method helps in pinpointing the exact range of numbers that make the inequality true, providing a solid foundation for constructing solution sets accurately.