Problem 45 - College Algebra

Question

Solve each linear inequality in Exercises 27-48 and graph the solution set on a number line. Express the solution set using interval notation. $$\frac{x-4}{6} \geq \frac{x-2}{9}+\frac{5}{18}$$

Step-by-Step Solution

Answer

The solution to the inequality is \(x \geq 13\) and the interval notation is \([13, \infty)\). The solution set on the number line extends from and includes the point 13 up to positive infinity.

Step 1: Simplify the inequality

First, it's important to simplify the inequality as much as possible. This can be done by collecting all terms including \(x\) on one side and constants on the other side. Then, multiply the entire inequality by 18 to get rid of denominators: \[18(\frac{x-4}{6}) \geq 18(\frac{x-2}{9})+18(\frac{5}{18})\] which simplifies to \[3x - 12 \geq 2x +1\]

Step 2: Isolate the variable

In order to isolate \(x\), subtract \(2x\) from both sides: \[3x - 2x \geq 2x - 2x +1 +12\] This simplifies to \[ x \geq 13\]

Step 3: Express the solution in interval notation

The inequality \(x \geq 13\) means that \(x\) includes 13 and extends to infinity. This can be represented in interval notation as: \[ [13, \infty) \]

Step 4: Graph the solution set on a number line

Start by drawing a number line, and place a closed circle at 13 (because the solution includes 13). Then, draw an arrow extending to the right to represent up to positive infinity

Key Concepts

Linear Inequality Graphing

Graphing linear inequalities is a way to visually represent the solution set of an inequality. It helps us see which values of the variable make the inequality true. In our example, we have the inequality \( x \geq 13 \). To graph this on a number line, we start by drawing a horizontal line, which is our number line. We then locate the number 13 on this line.

Since the inequality is \( \geq \) (greater than or equal to), we place a closed dot on 13, indicating that 13 is included in our set of solutions. From this point, we draw a thick line or an arrow to the right, signifying that all numbers greater than 13 are also solutions. Had the inequality been strict (\( > \) instead of \( \geq \)), we would use an open dot instead, signifying that 13 is not a part of the solution set. The graph is a visual representation that makes the range of solutions clear and unmistakable.

Interval Notation

Interval notation is a shorthand way to describe a range of numbers that are solutions to an inequality. It is particularly useful in expressing infinite ranges and is more concise than writing out all the numbers that satisfy the inequality. In interval notation, we use brackets [ ] to indicate that an endpoint is included in the set, and parentheses ( ) to show that an endpoint is not included.

For our example, the solution \( x \geq 13 \) is expressed in interval notation as \( [13, \infty) \). This tells us that our set includes the number 13 and all numbers greater than 13, extending onward towards infinity. The square bracket at 13 signifies inclusion, while the parenthesis at infinity indicates that infinity is not a specific number we can reach, but rather it describes a boundless extension.

Inequality Simplification

Simplifying an inequality involves manipulating it to get the variable of interest by itself on one side of the inequality, usually with a goal of making the inequality easier to solve or understand. In the given exercise, to simplify the inequality \( \frac{x-4}{6} \geq \frac{x-2}{9}+\frac{5}{18} \), we first clear the fractions by multiplying the entire inequality by the least common denominator, which in this case is 18. This results in a much simpler form, \( 3x - 12 \geq 2x + 1 \).

We can then isolate the variable by performing basic arithmetic operations. By subtracting \( 2x \) from both sides, we are left with \( x \geq 13 \). Simplification is a fundamental step that turns complex-looking inequalities into more tangible, solvable forms. Understanding each step of this process is crucial for solving linear inequalities correctly and is a skill that will be valuable in various mathematical contexts.