Problem 83 - Intermediate Algebra

Question

Solve the inequality in part a. Graph the solution set and write it in interval notation. Then use your answer to part a to determine the solution set for the inequality in part b. (No new work is necessary!) Graph the solution set and write it in interval notation. a. \(3(2 x+2)>5(x-1)+3 x\) b. \(3(2 x+2)<5(x-1)+3 x\)

Step-by-Step Solution

Answer

Part a: Solution set is \((-\infty, \frac{11}{2})\), Part b: Solution set is \([\frac{11}{2}, \infty)\).

Step 1: Simplify the Inequality in Part a

First, distribute the numbers inside the brackets: \(3 \cdot (2x + 2) = 6x + 6\) and \(5(x - 1) = 5x - 5\). Substitute these into the inequality: \(6x + 6 > 5x - 5 + 3x\).

Step 2: Combine Like Terms and Isolate x

Combine terms on the right: \(5x + 3x = 8x\). The inequality becomes: \(6x + 6 > 8x - 5\). Subtract \(6x\) from both sides to get \(6 > 2x - 5\). Then add \(5\) to both sides: \(11 > 2x\).

Step 3: Solve for x

Divide both sides by \(2\) to solve for \(x\): \(x < \frac{11}{2}\). Therefore, the solution set for part a in interval notation is \((-\infty, \frac{11}{2})\).

Step 4: Graph the Solution Set for Part a

Draw a number line and shade all values that are less than \(\frac{11}{2}\). We use an open circle at \(\frac{11}{2}\) to indicate it is not included in the solution set.

Step 5: Interpret Inequality for Part b

Part b is simply the reverse of part a, \(3(2x + 2) < 5(x - 1) + 3x\), meaning the solution set is the complement of the solution in part a. Thus, the solution is \([\frac{11}{2}, \infty)\).

Step 6: Graph the Solution Set for Part b

On a number line, shade all values greater than \(\frac{11}{2}\). Use an open circle at \(\frac{11}{2}\) since it is not part of the solution.

Key Concepts

Interval Notation

Interval notation is a method used to represent a set of numbers along the real number line. It is expressed in terms of the smallest and largest numbers in the set, addressing whether these bounds are included or not. Here's how you can use interval notation:

• Parentheses \(()\) are used to indicate that an endpoint is not included in the interval. They represent an open interval.
• Brackets \([])\) are used when an endpoint is included, indicating a closed interval.

Let's take an example from the solution where the solution set for part a is \((\infty, \frac{11}{2})\). This indicates that all numbers less than \(\frac{11}{2}\) are part of the solution, but \(\frac{11}{2}\) itself is not included. Similarly, in part b, the solution \([\frac{11}{2}, \infty)\) suggests that all numbers greater than \(\frac{11}{2}\) are included, but not \(\frac{11}{2}\) itself. This technique efficiently represents sets of solutions for inequalities.

Graphing Inequalities

Graphing inequalities involves representing the solution set of an inequality on a number line. This visual method helps in understanding which values satisfy the inequality. Here’s how to graph inequalities:

• Draw a number line for the range of interest.
• Identify the critical point (e.g., \(\frac{11}{2}\)) and mark it with an open circle to show it is not included (which represents that it's a strict inequality, \(<\) or \(>\)).
• If using a closed interval (like \(\leq\) or \(\geq\)), use a closed circle instead.
• Shade the relevant portion of the line where the inequality holds true.

In our exercise example:- For part a, shade all numbers to the left of \(\frac{11}{2}\) to represent all values less than \(\frac{11}{2}\).- For part b, shade all numbers to the right of \(\frac{11}{2}\) to illustrate values greater than \(\frac{11}{2}\).This approach allows you to visually verify your solution and ensure you have considered the inequality's direction correctly.

Solving Linear Inequalities

Solving linear inequalities is similar to solving linear equations, but with special attention to how the inequality sign can change. This process helps in finding the range of values that satisfy the inequality:

• Start by simplifying both sides, if necessary, through distribution or combining like terms.
• Isolate the variable by performing inverse operations, like addition or subtraction, just like you would in an equation.
• Remember that multiplying or dividing both sides by a negative number will reverse the inequality sign.

For example, let's revisit our original problem: 1. Start with simplifying the inequality \(3(2x + 2) > 5(x - 1) + 3x\) through distribution.2. Combine like terms to streamline the inequality.3. Isolate \(x\) by moving terms around.4. Ensure division by a positive number when solving for \(x\) to maintain the inequality direction.By following these steps, you determine that \(x < \frac{11}{2}\) is the solution for part a. The alteration of the inequality sign is crucial, so practice these techniques to boost confidence in handling similar problems.