Problem 59 - Algebra and Trigonometry

Question

Solve each rational inequality in Exercises \(43-60\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x-2}{x+2} \leq 2 $$

Step-by-Step Solution

Answer

The solution to the inequality is \((-∞, -2) \cup (-6, ∞)\)

Step 1: Simplify the rational inequality

In order to simplify the rational inequality \(\frac{x-2}{x+2} \leq 2\), subtract 2 from both sides to get it in a more recognizable form and to make the right side into 0. Thus, the inequality becomes \(\frac{x-2}{x+2} - 2 \leq 0\). This can be further simplified to \(\frac{x-2 - 2(x+2)}{x+2} \leq 0\) which gives us \(\frac{-x-6}{x+2} \leq 0\).

Step 2: Find the zero points

To find the zero points isolate x on one side of the inequality. This gives x = -6 or x = -2.

Step 3: Test the intervals

The number line is split into intervals by the zeroes. So test each interval by taking a point from each interval and substituting it back into the simplified inequality. Let's pick points -7, -3 and 0 from the intervals (-∞, -2), (-2, -6) and (-6, ∞) respectively. For -7, the inequality becomes \(\frac{-(-7)-6}{-7+2} \lessapprox 0\). For -3, it becomes \(\frac{-(-3)-6}{-3+2} \lessapprox 0\), and for 0, it becomes \(\frac{-0 - 6}{0+2} \lessapprox 0\). The first and the third inequalities are true

Step 4: Express the solution in interval notation

The solution on the number line are the intervals (-∞, -2) and (-6, ∞). When expressed in interval notation it is: \((-∞, -2) \cup (-6, ∞)\) . Note that the parenthesis means the endpoints are not included in the solution set. If a bracket was used then it would mean the endpoint is included.