Example: Using Interval Notation to Express All Real Numbers Greater Than or Equal to a Number

Use interval notation to indicate all real numbers greater than or equal to $-2$.

Answer: Use a bracket on the left of $-2$ and parentheses after infinity: $\left[-2,\infty \right)$. The bracket indicates that $-2$ is included in the set with all real numbers greater than $-2$ to infinity.

Example: Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b

Write the interval expressing all real numbers less than or equal to $-1$ or greater than or equal to $1$.

Answer: We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at $-\infty$ and ends at $-1$, which is written as $\left(-\infty ,-1\right]$. The second interval must show all real numbers greater than or equal to $1$, which is written as $\left[1,\infty \right)$. However, we want to combine these two sets. We accomplish this by inserting the union symbol, $\cup$, between the two intervals.

Example: Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:

1. $x - 15<4$
2. $6\ge x - 1$
3. $x+7>9$

Answer: The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality. 1. $$\begin{array}{ll}x - 15<4\hfill & \hfill \\ x - 15+15<4+15 \hfill & \text{Add 15 to both sides.}\hfill \\ x<19\hfill & \hfill \end{array}$$ 2. $$\begin{array}{ll}6\ge x - 1\hfill & \hfill \\ 6+1\ge x - 1+1\hfill & \text{Add 1 to both sides}.\hfill \\ 7\ge x\hfill & \hfill \end{array}$$ 3. $$\begin{array}{ll}x+7>9\hfill & \hfill \\ x+7 - 7>9 - 7\hfill & \text{Subtract 7 from both sides}.\hfill \\ x>2\hfill & \hfill \end{array}$$

Example: Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

1. $3x<6$
2. $-2x - 1\ge 5$
3. $5-x>10$

Answer: 1. $$\begin{array}{l}3x<6\hfill \\ \frac{1}{3}\left(3x\right)<\left(6\right)\frac{1}{3}\hfill \\ x<2\hfill \end{array}$$ 2. $$\begin{array}{ll}-2x - 1\ge 5\hfill & \hfill \\ -2x\ge 6\hfill & \hfill \\ \left(-\frac{1}{2}\right)\left(-2x\right)\ge \left(6\right)\left(-\frac{1}{2}\right)\hfill & \text{Multiply by }-\frac{1}{2}.\hfill \\ x\le -3\hfill & \text{Reverse the inequality}.\hfill \end{array}$$ 3. $$\begin{array}{ll}5-x>10\hfill & \hfill \\ -x>5\hfill & \hfill \\ \left(-1\right)\left(-x\right)>\left(5\right)\left(-1\right)\hfill & \text{Multiply by }-1.\hfill \\ x<-5\hfill & \text{Reverse the inequality}.\hfill \end{array}$$

Example: Solving an Inequality Algebraically

Solve the inequality: $13 - 7x\ge 10x - 4$.

Answer: Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the interval $\left(-\infty ,1\right]$, or all real numbers less than and including 1.

Example: Solving an Inequality with Fractions

Solve the following inequality and write the answer in interval notation: $-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x$.

Answer: We begin solving in the same way we do when solving an equation.

The solution set is the interval $\left(-\infty ,\frac{15}{34}\right]$.

Example: Solving a Compound Inequality

Solve the compound inequality: $3\le 2x+2<6$.

Answer: The first method is to write two separate inequalities: $3\le 2x+2$ and $2x+2<6$. We solve them independently.

Then, we can rewrite the solution as a compound inequality, the same way the problem began. In interval notation, the solution is written as $\left[\frac{1}{2},2\right)$. The second method is to leave the compound inequality intact and perform solving procedures on the three parts at the same time. We get the same solution: $\left[\frac{1}{2},2\right)$.

Example: Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts: $3+x>7x - 2>5x - 10$.

Answer: Let's try the first method. Write two inequalities:

The solution set is $-4<x<\frac{5}{6}$ or in interval notation $\left(-4,\frac{5}{6}\right)$. Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right as they appear on a number line.

Example: Determining a Number within a Prescribed Distance

Describe all values $x$ within a distance of 4 from the number 5.

Answer: We want the distance between $x$ and 5 to be less than or equal to 4. We can draw a number line to represent the condition to be satisfied. The distance from $x$ to 5 can be represented using an absolute value symbol, $|x - 5|$. Write the values of $x$ that satisfy the condition as an absolute value inequality.

We need to write two inequalities as there are always two solutions to an absolute value equation. If the solution set is $x\le 9$ and $x\ge 1$, then the solution set is an interval including all real numbers between and including 1 and 9. So $|x - 5|\le 4$ is equivalent to $\left[1,9\right]$ in interval notation.

Example: Solving an Absolute Value Inequality

Solve $|x - 1|\le 3$.

Answer:

$\begin{array}{c}|x - 1|\le 3\hfill \\ \hfill \\ -3\le x - 1\le 3\hfill \\ \hfill \\ -2\le x\le 4\hfill \\ \hfill \\ \left[-2,4\right]\hfill \end{array}$

Example: Using a Graphical Approach to Solve Absolute Value Inequalities

Given the equation $y=-\frac{1}{2}|4x - 5|+3$, determine the x-values for which the y-values are negative.

Answer: We are trying to determine where $y<0$ which is when $-\frac{1}{2}|4x - 5|+3<0$. We begin by isolating the absolute value.

Next, we solve $|4x - 5|=6$. Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $x=-\frac{1}{4}$ and $x=\frac{11}{4}$ and that the graph opens downward.