Chapter 2: Problem 35
For Problems \(35-44\), solve each compound inequality and graph the solution sets. Express the solution sets in interval notation. \(x-2>-1 \quad\) and \(\quad x-2<1\)
Short Answer
Expert verified
The solution is \((1, 3)\).
Step by step solution
01
Solve the first inequality
Let's solve the inequality \(x - 2 > -1\). Add 2 to both sides to isolate \(x\):\[x - 2 + 2 > -1 + 2\]Now, we have:\[x > 1\]
02
Solve the second inequality
Now, solve the inequality \(x - 2 < 1\). Add 2 to both sides to isolate \(x\):\[x - 2 + 2 < 1 + 2\]Which simplifies to:\[x < 3\]
03
Combine the inequalities
Since we have \(x > 1\) and \(x < 3\), we combine these to find the solution for the compound inequality. Combining these, we get:\[1 < x < 3\]
04
Graph the solution set
To graph the solution set \(1 < x < 3\), draw a number line. Place open circles around 1 and 3 to indicate that these points are not included in the solution, and shade the line between these two points.
05
Express the solution in interval notation
The interval notation for \(1 < x < 3\) is \((1, 3)\)This denotes all numbers greater than 1 and less than 3, but not including 1 and 3.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Inequalities
Graphing inequalities involves illustrating the range of solutions on a number line. When we graph the inequality \(1 < x < 3\), we start by imagining a simple horizontal line that represents all possible values of \(x\). This line, called a number line, helps us identify which values solve the inequality.
For the inequality \(1 < x < 3\), we will:
For the inequality \(1 < x < 3\), we will:
- Locate the numbers 1 and 3 on the number line.
- Use open circles at 1 and 3 to show these endpoints are not included in the solution set. Open circles are used because the inequality does not allow equality, the solution only considers values strictly greater than 1 and less than 3.
- Shade the section of the number line between the two open circles to indicate all values greater than 1 and less than 3 are part of the solution.
Interval Notation
Interval notation is a shorthand way of writing the set of numbers that are solutions to an inequality. It simplifies communication by condensing an entire interval of numbers into a single expression. For our inequality \(1 < x < 3\), the solution set can be written in interval notation as \((1, 3)\).
Here’s how to interpret this notation:
Here’s how to interpret this notation:
- The parentheses \(()\) mean that the endpoints are not included in the interval. In our example, this tells us that 1 and 3 themselves are not solutions, which aligns with the open circles used in graphing.
- If the interval included the endpoints, square brackets \([]\) would be used. For example, if our inequality was \(1 \leq x \leq 3\), the interval notation would be \([1, 3]\), indicating that 1 and 3 are included.
- Interval notation is efficient for representing the range of x values that solve the problem.
Inequality Solving Steps
Solving compound inequalities involves a step-by-step approach to finding the range of values that satisfy both parts of the inequality. Let's break down the solution steps for the inequality \(x-2>-1\) and \(x-2<1\).
Following these steps ensures the solution makes sense and fits the original problem accurately.
Step 1: Solve Each Inequality Separately
Start by isolating \(x\) in both inequalities:- For \(x - 2 > -1\), add 2 to both sides to simplify to \(x > 1\).
- For \(x - 2 < 1\), similarly, add 2 to both sides to get \(x < 3\).
Step 2: Combine the Solutions
To find a common solution that fits both inequalities, overlap the results from step 1. Combining \(x > 1\) and \(x < 3\) gives us the compound inequality \(1 < x < 3\).Step 3: Verify Through Graphing and Notation
Always graph the solution on a number line and express it in interval notation. For this problem, we graph \(1 < x < 3\) with open circles and shade between them, and write the interval as \((1, 3)\).Following these steps ensures the solution makes sense and fits the original problem accurately.