Problem 68 - Intermediate Algebra

Question

Solve each compound inequality, if possible. Graph the solution set (if one exists) and write it using interval notation. \(\frac{a}{2}+\frac{7}{4}>5\) or \(\frac{3}{8}+\frac{a}{3} \leq \frac{5}{12}\)

Step-by-Step Solution

Answer

The solution set is \((-\infty, \frac{1}{8}] \cup (6.5, \infty)\).

Step 1: Understanding the First Inequality

The first inequality is \( \frac{a}{2} + \frac{7}{4} > 5 \). To solve this inequality, we need to isolate \( a \).

Step 2: Solving the First Inequality

Subtract \( \frac{7}{4} \) from both sides: \( \frac{a}{2} > 5 - \frac{7}{4} \). This simplifies to \( \frac{a}{2} > \frac{20}{4} - \frac{7}{4} \), which further simplifies to \( \frac{a}{2} > \frac{13}{4} \). Multiply both sides by 2: \( a > \frac{13}{2} \) or \( a > 6.5 \).

Step 3: Understanding the Second Inequality

The second inequality is \( \frac{3}{8} + \frac{a}{3} \leq \frac{5}{12} \). We need to solve this inequality by isolating \( a \).

Step 4: Solving the Second Inequality

Subtract \( \frac{3}{8} \) from both sides: \( \frac{a}{3} \leq \frac{5}{12} - \frac{3}{8} \). To subtract these fractions, find a common denominator. The least common denominator is 24, so this becomes \( \frac{10}{24} - \frac{9}{24} \), which is \( \frac{1}{24} \). Thus, \( \frac{a}{3} \leq \frac{1}{24} \). Multiply both sides by 3: \( a \leq \frac{1}{8} \).

Step 5: Combining the Solutions

We have the separate solutions: \( a > 6.5 \) and \( a \leq \frac{1}{8} \). In a compound inequality combined with "or," the solution set is the union of the two separate inequalities. Since the first inequality represents numbers greater than 6.5, and the second represents numbers less than or equal to \( \frac{1}{8} \), these two sets do not overlap.

Step 6: Graphing the Solution Set and Writing Interval Notation

Since the two solution intervals do not overlap, the solution set is disjoint. On a number line, this means that the intervals are not connected. The final solution using interval notation is \( (-\infty, \frac{1}{8}] \cup (6.5, \infty) \).

Key Concepts

Solving Inequalities

When solving inequalities, the process is similar to solving equations, but with a key difference: you're dealing with inequalities instead of equalities. In the case of this exercise, we have compound inequalities, meaning that more than one inequality needs to be solved separately, and their solutions are then combined.
For the first inequality, \( \frac{a}{2} + \frac{7}{4} > 5 \), we want to isolate the variable \( a \). First, subtract \( \frac{7}{4} \) from both sides to get \( \frac{a}{2} > \frac{13}{4} \). Then, multiply both sides by 2, resulting in \( a > 6.5 \).
The second inequality is \( \frac{3}{8} + \frac{a}{3} \leq \frac{5}{12} \). Start by subtracting \( \frac{3}{8} \) from both sides, giving \( \frac{a}{3} \leq \frac{1}{24} \). After multiplying both sides by 3, you find \( a \leq \frac{1}{8} \).
Remember, when solving inequalities, pay close attention to the direction of the inequality sign. If you multiply or divide an inequality by a negative number, the direction of the inequality sign must be reversed. In this exercise, such a scenario did not occur.

Interval Notation

Interval notation is a mathematical method to describe a set of numbers as a pair of numbers. These numbers represent the endpoints of the interval.
Understanding the solution set of compound inequalities is made easier by using interval notation. In this exercise, we have two separate intervals: one from the inequality \( a > 6.5 \) and one from \( a \leq \frac{1}{8} \).
In interval notation:

• \( a > 6.5 \) is written as \( (6.5, \infty) \), indicating all numbers greater than 6.5 extending towards infinity.
• \( a \leq \frac{1}{8} \) is symbolized as \( (-\infty, \frac{1}{8}] \), indicating all numbers less than or equal to \( \frac{1}{8} \).

Each part of the compound inequality is written separately. The overall solution is the union of these intervals, combined using the symbol \( \cup \). Our result is \( (-\infty, \frac{1}{8}] \cup (6.5, \infty) \), which signifies that both intervals are part of the solution but don't overlap.

Graphing Inequalities

Graphing inequalities involve representing the solutions on a number line to visually communicate the solution set.
For the compound inequality in this exercise, we plot two separate intervals:

• For \( a > 6.5 \), draw an open circle at 6.5 on the number line and shade everything to the right, extending indefinitely. An open circle indicates that 6.5 is not included in the set.
• For \( a \leq \frac{1}{8} \), place a closed circle at \( \frac{1}{8} \) and shade all values to the left, as low as negative infinity. A closed circle denotes that \( \frac{1}{8} \) is included in the set.

The graph visually confirms that there is no overlap in the intervals, just like the interval notation suggests. This helps in visually verifying the solution, ensuring clarity and understanding.