Problem 55 - Elementary and Intermediate Algebra

Question

Solve each compound inequality. Graph the solution set and write it using interval notation. $$ 2 a+10<7 a \text { and } 5 a-15<2 a $$

Step-by-Step Solution

Answer

The solution is \(2 < a < 5\) or \((2, 5)\) in interval notation.

Step 1: Solve the First Inequality

The first inequality is \(2a + 10 < 7a\). We start by isolating \(a\) on one side. Subtract \(2a\) from both sides to get \(10 < 5a\). Then, divide both sides by 5 to solve for \(a\), giving us \(a > 2\).

Step 2: Solve the Second Inequality

The second inequality is \(5a - 15 < 2a\). Start by subtracting \(2a\) from both sides to get \(3a - 15 < 0\). Then, add 15 to both sides, which gives \(3a < 15\). Finally, divide both sides by 3 to solve for \(a\), resulting in \(a < 5\).

Step 3: Combine the Inequalities

The solution to the compound inequality is the intersection of both solutions from the previous steps. This gives \(2 < a < 5\).

Step 4: Write in Interval Notation

The interval notation representation of the solution \(2 < a < 5\) is \((2, 5)\).

Step 5: Graph the Solution Interval

To graph the solution set \((2, 5)\), draw a number line, highlight the region between 2 and 5, and use open circles at both 2 and 5 to indicate that these values are not included in the solution.

Key Concepts

Graphing Inequalities

Graphing inequalities involves visually representing the solutions of inequalities on a number line. This practise helps in clearly understanding the range of values that satisfy a given inequality. To graph the compound inequality \(2 < a < 5\), begin by drawing a horizontal number line with reference points labeled, especially around the numbers in the inequality.

• Mark the number 2 and 5 on the number line. These numbers are boundaries of the solution set.
• Use open circles above 2 and 5 to show that these values are not included in the solution, as they are part of strict inequalities (\(<\)).
• Shade or highlight the region between these two points to represent the values \(a\) can take (any value greater than 2 and less than 5).

By doing this, you provide a clear visual indication of the segment of the number line that contains all possible solutions. Graphing is particularly useful when dealing with compound inequalities, as it makes it easier to see the intersection of different solution sets.

Interval Notation

Interval notation is a concise and effective way to represent a range of solutions. In interval notation, parentheses and brackets are used to describe the endpoints of an interval on a number line.

• Parentheses \((\) indicate that an endpoint is not included in the interval; this is known as an open interval.
• Brackets \([\) show that an endpoint is included, which is referred to as a closed interval.

For the compound inequality solution \(2 < a < 5\), which indicates \(a\) is greater than 2 and less than 5, use open parentheses because 2 and 5 themselves are not part of the solution. Thus, in interval notation, this solution would be written as \((2, 5)\). This succinct form communicates the same idea as the inequality, but in a way that is often quicker to read and understand in the context of solving math problems.

Solving Inequalities

Solving inequalities is akin to solving equations, but with careful consideration of the direction of the inequality. In a compound inequality, you're tasked with finding values that satisfy multiple conditions simultaneously.To solve the first inequality \(2a + 10 < 7a\):

• Subtract \(2a\) from both sides, simplifying the expression to \(10 < 5a\).
• Divide both sides by 5 to isolate \(a\), leading to \(a > 2\).

Similarly, for the second inequality \(5a - 15 < 2a\):

• Subtract \(2a\) from both sides to get \(3a - 15 < 0\).
• Add 15 to both sides, yielding \(3a < 15\).
• Finally, divide by 3 to solve for \(a\), which gives \(a < 5\).

Combine these two inequalities to form a compound inequality: \(2 < a < 5\). Remember that solving inequalities requires maintaining the inequality's direction (keep an eye on when you multiply or divide by a negative, as it reverses the inequality sign—which wasn't the case here). This solution set represents the values that satisfy both conditions at once.