Problem 51 - Beginning and Intermediate Algebra

Question

Solve each rational inequality. Graph the solution set and write the solution in interval notation. $$\frac{7}{t+6}<3$$

Step-by-Step Solution

Answer

The solution to the rational inequality is \((-\infty, -6) \cup (-6, -\frac{11}{3})\).

Step 1: Isolate the rational expression

Subtract 3 from both sides of the inequality: \(\frac{7}{t+6} - 3 < 0\)

Step 2: Find the undefined values

The rational expression is undefined when the denominator is equal to 0: \(t + 6 = 0\) \[t = -6\]

Step 3: Set the inequality equal to zero and solve for t

Set the inequality equal to zero and solve for t: \[\frac{7}{t+6} - 3 = 0\] \[\frac{7}{t+6} = 3\] \[7 = 3(t+6)\] \[7 = 3t + 18\] \[t = -\frac{11}{3}\]

Step 4: Construct a number line using the critical points

Using the values obtained in steps 2 and 3, construct a number line with the critical points -6 and -\(\frac{11}{3}\), with the following intervals: 1. t < -6 2. -6 < t < -\(\frac{11}{3}\) 3. t > -\(\frac{11}{3}\)

Step 5: Test each interval using the inequality

Test each interval of the number line using the original inequality. 1. \(t = -7\): \[\frac{7}{-7+6} - 3 < 0 \Rightarrow \frac{7}{-1} - 3 < 0\] 2. \(t = -4\): \[\frac{7}{-4+6} - 3 < 0 \Rightarrow \frac{7}{2} - 3 < 0\] 3. \(t = 0\): \[\frac{7}{0+6} - 3 < 0 \Rightarrow \frac{7}{6} - 3 < 0\]

Step 6: Graph the solution set

The intervals that satisfy the inequality are: 1. t < -6 (For example, t = -7) 2. -6 < t < -\(\frac{11}{3}\) (For example, t = -4) So we can graph the solution set on the number line, with open circles over -6 and -\(\frac{11}{3}\) to indicate that these values are not included in the solution.

Step 7: Write the solution in interval notation

Using the intervals we determined, the solution in interval notation is: \((-\infty, -6) \cup (-6, -\frac{11}{3})\)

Key Concepts

Understanding the Solution Set

A "solution set" refers to all the possible values that satisfy a given inequality. When solving rational inequalities, like \(\frac{7}{t+6} < 3\), we first find values where the expression isn't defined. Here, the inequality is undefined at \(t = -6\), since division by zero is impossible.
Next, we solve \(\frac{7}{t+6} = 3\) to find critical points, giving us \(t = -\frac{11}{3}\). These critical points help us divide the number line into test intervals. By evaluating the inequality within these intervals, we find those that satisfy the solution, forming the solution set.
The solution set for this inequality includes values for which the expression holds true, and is expressed using open circles to show excluded endpoints like \(-6\) and \(-\frac{11}{3}\). This means those values are not part of the solution set.

Expressing the Solution in Interval Notation

Interval notation is a shorthand way of writing the set of solutions that satisfy an inequality. For the inequality \(\frac{7}{t+6} < 3\), we use critical points to divide the number line. The suitable intervals here are \(t < -6\) and \(-6 < t < -\frac{11}{3}\).
This is expressed in interval notation as \((-finity, -6) \cup (-6, -\frac{11}{3})\).

• The round brackets \(()\) indicate that the endpoints are not included in the solution set, which is consistent with open circles in graphing.
• The union symbol \(\cup\) is used to combine intervals that are part of the solution.
• Remember, \(-\infty\) with \(\infty\) always uses round brackets because infinity isn't a specific number but a concept.

Interval notation is efficient for expressing ranges and is used frequently in mathematics.

Graphing Inequalities on a Number Line

Graphing inequalities helps visually represent the solution set on a number line. For our inequality \(\frac{7}{t+6} < 3\), we graph the solution set determined in previous steps.
We use open circles at \(-6\) and \(-\frac{11}{3}\) to indicate these points are not included, then shade the line on both sides of these circles:

• One shaded part should go leftward from \(-6\) towards \(-\infty\).
• Another shaded section runs between \(-6\) and \(-\frac{11}{3}\).

These shaded regions show where the inequality holds true, providing a clear visual distinction of the solution set in relation to the entire number line. Graphing this solution allows you to easily see which intervals of \(t\) fulfill the inequality condition.