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Chapter 3: Problem 55

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. $$ \frac{x+1}{x+3}<2 $$

Short Answer

Expert verified
The solution set for the given inequality in interval notation is \((-3, \infty)\).

Step by step solution

01

Isolate the Variable on One Side

Write the inequality to isolate the variable on one side: \( \frac{x+1}{x+3} - 2 < 0 \). Simplify this: \( \frac{x+1-2(x+3)}{x+3} < 0 \). The equation becomes \( \frac{-x-5}{x+3} < 0 \).
02

Transform into Equivalent Inequalities

To eliminate the denominator, transform the inequality into a set of equivalent inequalities. The parts are being separated by the roots of the equation \(-x-5 = 0\) and \(x+3 = 0\). These are \(x = -5\) and \(x = -3\). The x number line is divided into three intervals by these roots.
03

Sign Analyses on Each of the Intervals

Examine the signs in each of the intervals using the roots -5 and -3. Consider the intervals \((- \infty, -5) \), \((-5, -3) \) and \((-3, \infty) \). In the first interval when x<-5, the expression \(-x-5 > 0\) and \(x+3 < 0\). The overall result is positive as the signs are opposite. In the second interval when -5-3, the expression \(-x-5 < 0\) and \(x+3 > 0\). The overall result is negative as the signs are opposite.
04

Identify Solution Interval

For the inequality \( \frac{-x-5}{x+3} < 0 \) to hold true, the expression needs to be negative. This occurs for \(x>-3\). Thus the required solution for original inequality is \(( -3, \infty) \). However, the root \(x=-3\) also needs to be removed from the solution set since it would make the denominator of the fraction zero, and division by zero is undefined.

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