Problem 30 - College Algebra with Modeling and Visualization

Question

Solve the inequality symbolically. Express the solution set in set-builder or interval notation. $$ \frac{8}{3} \geq \frac{4}{3}-(x+3) \geq \frac{2}{3} $$

Step-by-Step Solution

Answer

The solution is \([-\frac{13}{3}, -\frac{7}{3}]\).

Step 1: Break Down Compound Inequality

The given compound inequality is \( \frac{8}{3} \geq \frac{4}{3}-(x+3) \geq \frac{2}{3} \). This is equivalent to two separate inequalities: \( \frac{8}{3} \geq \frac{4}{3} - (x+3) \) and \( \frac{4}{3} - (x+3) \geq \frac{2}{3} \). We'll solve each one separately.

Step 2: Solve the First Inequality

Start with the first inequality: \( \frac{8}{3} \geq \frac{4}{3} - (x+3) \).Simplify the expression on the right: \( \frac{4}{3} - (x+3) = \frac{4}{3} - x - 3 = \frac{4}{3} - \frac{9}{3} - x = -\frac{5}{3} - x \).Now, the inequality is \( \frac{8}{3} \geq -\frac{5}{3} - x \).Add \( x \) to both sides: \( \frac{8}{3} + x \geq -\frac{5}{3} \).Add \( \frac{5}{3} \) to both sides: \( \frac{8}{3} + \frac{5}{3} + x \geq 0 \).Combine terms: \( \frac{13}{3} + x \geq 0 \).Thus, \( x \geq -\frac{13}{3} \).

Step 3: Solve the Second Inequality

Now solve the second inequality: \( \frac{4}{3} - (x+3) \geq \frac{2}{3} \).Simplify the expression on the left: \( \frac{4}{3} - x - 3 \) becomes \( -x - \frac{5}{3} \).The inequality is now \( -x - \frac{5}{3} \geq \frac{2}{3} \).Add \( x \) to both sides: \( - \frac{5}{3} \geq \frac{2}{3} + x \).Add \( \frac{5}{3} \) to both sides: \( 0 \geq \frac{7}{3} + x \).Subtract \( \frac{7}{3} \) from both sides: \( -\frac{7}{3} \geq x \), which simplifies to \( x \leq -\frac{7}{3} \).

Step 4: Find the Solution Set

The solution to the compound inequality consists of both solutions: \( x \geq -\frac{13}{3} \) and \( x \leq -\frac{7}{3} \). The overlap, or intersection, of these two solutions is: \( x \) must satisfy \( -\frac{13}{3} \leq x \leq -\frac{7}{3} \).Therefore, the solution set in interval notation is \( \left[ -\frac{13}{3}, -\frac{7}{3} \right] \).

Key Concepts

Compound Inequalities

Compound inequalities involve two separate inequalities joined together by either "and" or "or." In this exercise, we are dealing with a compound inequality that uses the word "and." This means both inequalities must be true at the same time.
Imagine having two inequalities combined into one statement, as we do with \(\frac{8}{3} \geq \frac{4}{3}-(x+3) \geq \frac{2}{3}\). What we do is break this down into two separate parts to solve individually.

• First, handle \(\frac{8}{3} \geq \frac{4}{3}-(x+3)\).
• Then solve \(\frac{4}{3}-(x+3) \geq \frac{2}{3}\).

Only after solving the separate inequalities do we find the intersection of their solutions. This intersection is your answer for the compound inequality, stating where both conditions overlap.

Interval Notation

Interval notation is a way of writing subsets of the real number line. It’s a neat shorthand method that lets you convey your solution succinctly. In interval notation, an interval consists of two numbers representing the bounds of the solution set.
For closed intervals, which include the endpoints, square brackets \([\text{ and } ]\) are used. Open intervals use parentheses \((\text{ and } )\), indicating that the endpoints are not included in the set. In our case, we have the solution \(x \geq -\frac{13}{3}\) and \(x \leq -\frac{7}{3}\).
Once both parts are solved, you combine them into a single interval that shows where the solutions overlap. Here, the overlap is shown as a closed interval: \(-\frac{13}{3} \leq x \leq -\frac{7}{3}\). In interval notation, this becomes \([-\frac{13}{3}, -\frac{7}{3}]\). This tells us the solution includes both endpoints.

Set-Builder Notation

Set-builder notation is another way to specify a set of solutions, and it provides more detail about what elements belong in your set. It uses curly braces \(\{\text{ and }\}\) and provides a condition for the elements of the set.
This form can be more descriptive, especially when explaining complex conditions. For example, in this exercise the solution \(-\frac{13}{3} \leq x \leq -\frac{7}{3}\) can be expressed in set-builder notation as:

• \(\{x \mid -\frac{13}{3} \leq x \leq -\frac{7}{3}\}\)

This reads as "the set of all \(x\) such that \(x\) is greater than or equal to \(-\frac{13}{3}\) and less than or equal to \(-\frac{7}{3}\)." This approach not only tells you the bounds but also emphasizes the conditions that must be satisfied for each element in the set.