Problem 14 - Algebra for College Students

Question

Solve each inequality and graph its solution set on a number line. $$\frac{x-1}{x+2}>0$$

Step-by-Step Solution

Answer

The solution is \((-\infty, -2) \cup (1, \infty)\) on a number line.

Step 1: Find the Critical Points

Determine where the inequality changes sign by finding when the expression is zero or undefined. Set the numerator \(x - 1 = 0\) which gives \(x = 1\). Set the denominator \(x + 2 = 0\) which gives \(x = -2\). These points divide the number line into intervals.

Step 2: Test the Intervals Using the Critical Points

Select test points from the intervals created by the critical points to see where the inequality holds. We have the intervals \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\). Choose a point in each interval: - Test point \(x = -3\) in \((-\infty, -2)\): \(\frac{-3-1}{-3+2} = \frac{-4}{-1} = 4\) which is positive.- Test point \(x = 0\) in \((-2, 1)\): \(\frac{0-1}{0+2} = \frac{-1}{2}\) which is negative.- Test point \(x = 2\) in \((1, \infty)\): \(\frac{2-1}{2+2} = \frac{1}{4}\) which is positive.

Step 3: Determine the Solution Set

The inequality \(\frac{x-1}{x+2} > 0\) holds in the intervals where the expression is positive. From Step 2, this is true for intervals \((-\infty, -2)\) and \((1, \infty)\). So the solution set is \((-\infty, -2) \cup (1, \infty)\).

Step 4: Graph the Solution on a Number Line

Mark the critical points \(-2\) and \(1\) on the number line. Use open circles at these points to indicate they are not included in the solution. Shade the regions to the left of \(-2\) and to the right of \(1\) to represent the solution set \((-\infty, -2) \cup (1, \infty)\).

Key Concepts

Critical Points

When solving inequalities like \(\frac{x-1}{x+2}>0\), identifying critical points is the first crucial step. Critical points are values of \(x\) where the expression might change its sign. These can be points where the expression is either undefined or equals zero.

• The numerator sets up the equation \(x-1=0\). Solving gives \(x = 1\).
• The denominator leads to \(x+2=0\), which gives \(x = -2\).

These critical points \(-2\) and \(1\) divide the number line into different segments or intervals. By determining these points, you set the backdrop for testing where the inequality holds true or false.

Interval Testing

Once you have your critical points, the next step is interval testing. This involves choosing test points within the identified intervals to examine if they satisfy the inequality \(\frac{x-1}{x+2}>0\). By checking a single point in each interval, you can determine the sign (positive or negative) of the expression throughout that interval.

• For interval \((-\infty, -2)\), choose \(x=-3\). This gives \(\frac{-4}{-1}=4\), which is positive.
• In interval \((-2, 1)\), choose \(x=0\). This results in \(\frac{-1}{2}\), a negative value.
• Lastly, for \((1, \infty)\), \(x=2\) gives \(\frac{1}{4}\), which is positive.

Through this method, you discover where the inequality holds, aiding in identifying the solution set.

Solution Set

The solution set includes all values of \(x\) where the inequality \(\frac{x-1}{x+2}>0\) is true. From interval testing, you found positive values in intervals \((-\infty, -2)\) and \((1, \infty)\).
To construct the solution set:

• The interval \((-\infty, -2)\) is valid because the test point was positive, indicating all values in this region satisfy the inequality.
• Similarly, \((1, \infty)\) gives positive results and is also part of the solution.

Therefore, the solution set is \((-\infty, -2) \cup (1, \infty)\). Remember, these intervals use open brackets because the inequality is strict (\(> 0\) rather than \(\geq 0\)).

Graphing Inequalities

Graphing inequalities visually represents the solution set on a number line, providing a clear image of which parts of the line satisfy the inequality. Here’s how it’s done:

• Mark the critical points \(-2\) and \(1\) on a number line.
• Place open circles at these points to show that they are not part of the solution (since the inequality is strict).
• Shade the regions stretching from \(-\infty\) to \(-2\), and from \(1\) to \(+\infty\), indicating the solution set \((-\infty, -2) \cup (1, \infty)\).

The visual representation makes it easy to comprehend where the expression is positive and which values of \(x\) are included in the solution.