Problem 80 - Intermediate Algebra

Question

Solve the inequality in part a. Graph the solution set and write it in interval notation. Then use your work from part a to determine the solution set for the compound inequality in part b. (No new work is necessary!) Graph the solution set and write it in interval notation. a. \(7 \leq 4 x+1<23\) b. \(7<4 x+1 \leq 23\)

Step-by-Step Solution

Answer

Part a: \([\frac{3}{2}, 5.5)\); Part b: \((\frac{3}{2}, 5.5]\).

Step 1: Break Down the Inequality for Part a

The inequality to solve is \(7 \leq 4x + 1 < 23\). This is a compound inequality which means it involves two separate inequalities: \(7 \leq 4x + 1\) and \(4x + 1 < 23\). We will solve each part independently.

Step 2: Solve the Left Part of the Compound Inequality

For the inequality \(7 \leq 4x + 1\), subtract 1 from both sides to isolate the term with \(x\). This results in \(6 \leq 4x\). Now, divide both sides by 4 to solve for \(x\): \(\frac{6}{4} \leq x\), which simplifies to \(\frac{3}{2} \leq x\).

Step 3: Solve the Right Part of the Compound Inequality

For the inequality \(4x + 1 < 23\), subtract 1 from both sides, giving \(4x < 22\). Divide both sides by 4 to find \(x\), giving \(x < \frac{22}{4}\), which simplifies to \(x < 5.5\).

Step 4: Combine the Solutions from Part a

Combine the results from Step 2 and Step 3. The solution for \(7 \leq 4x + 1 < 23\) is \(\frac{3}{2} \leq x < 5.5\).

Step 5: Graph the Solution and Write in Interval Notation for Part a

Graph the interval \([\frac{3}{2}, 5.5)\) on a number line. This includes the number \(\frac{3}{2}\) and extends to, but does not include, the number 5.5. Represent this interval in interval notation as \([\frac{3}{2}, 5.5)\).

Step 6: Determine the Solution for Part b Using Part a

Since we previously solved \(7 \leq 4x + 1 < 23\), for the inequality \(7 < 4 x + 1 \leq 23\), the solution will be similar. Drop the equality from the left endpoint and add it to the right endpoint. Therefore, the solution for part b is \((\frac{3}{2}, 5.5]\).

Step 7: Graph the Solution and Write in Interval Notation for Part b

Graph the interval \((\frac{3}{2}, 5.5]\) on a number line. This does not include the number \(\frac{3}{2}\), but it includes the number 5.5. Write this interval in interval notation as \((\frac{3}{2}, 5.5]\).

Key Concepts

Graphing Inequalities

Graphing inequalities is an essential skill in understanding how solutions to inequalities are represented. When you solve an inequality, the solution isn't just a single number but typically a range of numbers. These are expressed visually on a number line. Here’s how it works:

First, you'll need to find the boundary points of the inequality from its solution. For instance, in the inequality \([\frac{3}{2}, 5.5)\), the numbers \(\frac{3}{2}\) and \(5.5\) are the boundary points.

Next, use circles on these points to indicate whether they are included (closed circle) or not (open circle). \(\frac{3}{2}\) has a closed circle in \(\frac{3}{2} \leq x < 5.5\), and \(5.5\) has an open circle. However, in \((\frac{3}{2}, 5.5]\), \(\frac{3}{2}\) has an open circle, and \(5.5\) has a closed circle.

Finally, shade the section of the number line between the boundary points to show all the numbers that satisfy the inequality. This visual representation helps to clearly see which numbers are included in the solution.

Interval Notation

Interval notation is a concise way to express a range of numbers that are solutions to an inequality. It provides a neat and efficient way to display these solutions.

In interval notation, each interval is written between square or round brackets. A square bracket \[ \]\ signifies that the boundary number is included in the solution set (closed interval). A round bracket \( \)\ means the boundary number is excluded (open interval).

For example:

• \([\frac{3}{2}, 5.5)\) indicates that the numbers equal to \(\frac{3}{2}\) are included, but those equal to \(5.5\) aren't.
• \((\frac{3}{2}, 5.5]\) indicates that the numbers greater than \(\frac{3}{2}\) but less than or equal to \(5.5\) are solutions.

Using interval notation simplifies the representation of solutions and is widely used in mathematical discussions and solutions.

Solving Linear Inequalities

Solving linear inequalities is similar to solving linear equations but involves finding a range of solutions instead of a single solution. A linear inequality has variables, like a linear equation, but with inequality symbols such as < or \(\leq\).

To solve a linear inequality:

• Isolate the variable by performing arithmetic operations like addition, subtraction, multiplication, or division, ensuring you do the same on both sides of the inequality.
• Remember that multiplying or dividing by a negative number will flip the inequality sign.

For instance, in solving \(7 \leq 4x + 1 < 23\), split it into two separate inequalities, solve them, and combine the results. Start with \(7 \leq 4x + 1\): subtract 1, then divide by 4 to get \(x \geq\ \frac{3}{2}\). For \(4x + 1 < 23\), subtract 1 and divide by 4, leading to \(x < 5.5\).

Lastly, merge these results to get the complete solution in \(\frac{3}{2} \leq x < 5.5\). Solving inequalities accurately involves careful attention to these steps, understanding when to flip the inequality, and ensuring the range of solutions is captured correctly.