Survey of Algebra and Probability: MAT 101

Subsection 2.B.1 Intervals and Inequalities

When we solve an equation we find a single value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use an equals sign, but one of the following symbols:

The expression \(x \lt 4\) this means our variable \(x\) can be any number smaller than \(4\) such as \(-2, 0, 3, 3.9\) or even \(3.999999999\) as long as it is smaller than \(4\text{.}\) In other words, \(x \lt 4\) is the set of all numbers less than \(4\text{.}\) 4 is NOT less than 4. We write \(4\nless 4\text{.}\) However \(4\) IS less than or equal to \(4\text{.}\) We write \(4\leq 4\text{.}\)

The expression \(y \geq -2,\) means that the variable \(y\) can be any number greater than or equal to \(-2,\) such as \(5, 0,-1,-1.9999,\) or even \(-2\text{.}\) In other words, \(x \geq -2\) is the set of all numbers greater than or equal to \(-2\text{.}\)

It is often useful to draw a picture of the solutions to the inequality on a number line. We will start from the value in the problem and bold the lower part of the number line if the variable is smaller than the number, and bold the upper part of the number line if the variable is larger. The value itself we will mark with an open or closed circle: open for less than or greater than, and a closed circle for less than or equal to or greater than or equal to.

Once the graph is drawn we can quickly convert the graph into what is called interval notation. Interval notation gives two numbers, the first is the smallest value (furthest left on the number line), the second is the largest value (furthest right on the number line). We will use square brackets if the inequality includes or equal to (so either \(\leq\) or \(\geq\)). We will use round brackets if the inequality is strictly less than or greater than (so either \(\lt\) or \(\gt\)). If there is no largest value, we can use \(\infty\) (infinity). If there is no smallest value, we can use \(-\infty\) (negative infinity). If we use either positive or negative infinity we will always use a round bracket by the symbol.

Example 2.B.1. Relating an Inequality, Graph and Interval.

Graph the inequality \(x\geq 4\) and give the interval notation.

Start at \(4\) and shade to the right. Use a closed circle for greater than or equal to.

Our Graph\(\checkmark\)

Interval notation: \([4,\infty ) \checkmark\)

Example 2.B.2. Relating an Inequality, Graph and Interval.

Graph the inequality \(x\lt -4\) and give the interval notation.

Start at \(-4\) and shade to the left. Use an open circle for less than.

Our Graph\(\checkmark\)

Interval notation: \((-\infty, -4) \checkmark\)

Example 2.B.3. Relating an Inequality, Graph and Interval.

Graph the inequality \(-3\lt x \lt 1\) and give the interval notation.

Start at \(-3\) and shade to the right to \(1\text{.}\) Use open circles on both ends for less than.

Our Graph\(\checkmark\)

Interval notation: \((-3, -1) \checkmark\)

Subsection 2.B.2 Linear Inequalities

Solving inequalities is very similar to solving equations with one exception. To understand the exception, consider the tools we use to solve an equation: add/subtract, multiply/divide numbers to both sides of the equation to isolate the variable. We consider the inequality \(1 \lt 3\) and notice what happens to the inequality sign as we add, subtract, multiply and divide by both positive and negative numbers.

First we look at that happens when we add positive or negative numbers.

Consider \(1 \lt 3\text{.}\) Add \(3\) to both

to get \(4 \lt 6\text{.}\)

The result is a true statement.

Add \(-3\) to both

to get \(-2 \lt 0\text{.}\)

The result is still a true statement.

Note that adding \(-3\) and subtracting \(3\) produce the same result. Next we consider multiplication.

For \(1 \lt 3\text{,}\) multiply both by \(2\)

to get \(2 \lt 6\text{,}\) which is true.

This time multiply both by \(-2\text{.}\)

We get \(-2\) and \(-6\)

but \(-2\) is greater than \(-6\text{.}\)

The result is “flipped”,\(-2 \gt -6\text{.}\)

Let's try this again, but starting with a negative number: \(-3 \lt 1\)

For \(-3 \lt 1\text{,}\) multiply by \(-2\)

We get \(6\) and \(-2\)

and again the result switches the

inequality since \(6 \gt -2\text{.}\)

Divide both by \(2\text{.}\)

We get \(\frac{1}{2}\) and \(\frac{3}{2}\)

and \(\frac{1}{2} \lt \frac{3}{2}\text{.}\)

Let's try this again, but starting with a negative number: \(-3 \lt 1\)

This time divide by \(-2\)

to get \(\frac{3}{2}\) and \(-\frac{1}{2}.\)

The result switches the

inequality \(\frac{3}{2} \gt -\frac{1}{2}\text{.}\)

As the above examples illustrate, we can add, subtract, multiply, or divide on both sides of the inequality. But if we multiply or divide by a negative number, the inequality symbol will need to "flip" directions. We will keep that in mind as we solve inequalities.

Example 2.B.4. Solve an Inequality.

Solve and give the interval notation.

\begin{align*} 5-2x\amp\geq 11 \amp\amp \text{Subtract \(5\) from both sides}\\ \underline{-5\phantom{1234}}\amp\underline{\phantom{1}-5} \amp\amp \,\\ -2x\amp\geq 6 \amp\amp \text{Divide both sides by \(-2\)}\\ \overline{-2}\amp\phantom{12}\overline{-2} \amp\amp \text{Divide by a negative: flip inequality sign!}\\ x\amp\color{red}{\leq}\color{black}{}-3 \amp\amp \text{Graph, starting at \(-3\), going left with a solid circle for less than or equal to} \end{align*}

Our Graph\(\checkmark\)

Interval notation: \((-\infty, -3] \checkmark\)

The inequality we solve can get as complex as the linear equations we solved. We will use all the same patterns to solve these inequalities as we did for solving equations. Just remember that any time we multiply or divide by a negative number the inequality symbol switches directions (multiplying or dividing by a positive does not change the symbol!)

Example 2.B.6. Solve an Inequality.

Solve and give the interval notation.

\begin{align*} 3(2x-4)+4x\amp\lt 4(3x-7)+8 \amp\amp \text{Distribute}\\ 6x-12+4x\amp\lt 12x-28+8 \amp\amp \text{Combine like terms}\\ 10x-12\amp\lt 12x-20 \amp\amp \text{"Move" variable to one side}\\ \underline{-10x\phantom{1234}}\amp\underline{\phantom{1}-10x} \amp\amp \text{Subtract \(10x\) from both sides}\\ -12\amp\lt 2x-20 \amp\amp \text{Isolate variable on \textbf{RHS}}\\ \underline{+20}\amp\underline{\phantom{12345}+20} \amp\amp \text{Add \(20\) to both sides}\\ 8\amp\lt 2x \amp\amp \text{Divide both sides by \(2\)}\\ \overline{2}\amp\phantom{123}\overline{2}\amp\amp \text{Divide by a positive: DON'T flip inequality sign!}\\ 4\amp\lt x \amp\amp \text{Be careful with graph, \(x\) is larger!} \end{align*}

Our Graph\(\checkmark\)

Interval notation: \((4, \infty)\checkmark\)

It is important to be careful when the inequality is written backwards as in the previous example ( \(4 \lt x\) rather than \(x \gt 4\)). The inequality symbol opens to the variable, this means the variable is greater than \(4\text{.}\) So we must shade to the right of \(4\text{.}\)

Subsection 2.B.3 Compound Inequalities

Several inequalities can be combined together to form what are called compound inequalities. We will consider one type of compound inequality in this lesson.

Example 2.B.8. Three-part Compound Inequality.

Solve the compound inequality, graph the solution, and express it in both inequality and interval notation.

\(-4\leq 2x-4 \lt 2\)

The inequality requires the expression \(\color{blue}{2x-4}\) to lie in the interval \([-4,2)\)

\begin{align*} -4\leq 2x-4 \lt 2 \amp\amp\amp \text{Add \(4\) to all three parts of the compound inequality}\\ \underline{+4\phantom{1234}+4\phantom{1}+4} \amp\amp\amp \,\\ 0\leq 2x \lt 6 \amp\amp\amp \text{Divide all three parts by \(2\)}\\ \overline{ 2 }\phantom{123}\overline{ 2 }\phantom{1234}\overline{ 2 } \amp\amp\amp \, \end{align*}

\(0\leq \color{red}{x} \lt 3\)

\begin{align*} 0\leq x \lt 3 \amp\amp\amp \text{Our Solution: Inequality Notation}\checkmark\\ [0,3) \amp\amp\amp \text{Our Solution: Interval Notation}\checkmark \end{align*}

Example 2.B.9. Three-part Compound Inequality (Negative Coefficient).

Solve the compound inequality, graph the solution, and express it in both inequality and interval notation.

\(-3\lt 7-2x \leq -1\)

The inequality requires the expression \(\color{blue}{7-2x}\) to lie in the interval \((-3,-1]\)

\begin{align*} -3\lt 7-2x \leq -1 \amp\amp\amp \text{Subtract \(7\) from all three parts of the compound inequality}\\ \underline{-7\phantom{1}-7\phantom{1234}-7} \amp\amp\amp \,\\ -10\lt -2x \leq -8 \amp\amp\amp \text{Divide all three parts by \(-2\) (FLIP SIGNS)}\\ \overline{ -2 }\phantom{123}\overline{ -2 }\phantom{1234}\overline{ -2 } \amp\amp\amp \, \end{align*}

\(5\gt \color{red}{x}\geq 4\)

\begin{align*} 4\leq x \lt 5 \amp\amp\amp \text{Our Solution: Inequality Notation}\checkmark\\ [4,5) \amp\amp\amp \text{Our Solution: Interval Notation}\checkmark \end{align*}

Subsection 2.B.4 Absolute Value Inequalities

We now turn our attention to absolute value inequalities. Inequalities involving absolute values are used in statistics and other fiields. Consider

Absolute value is defined as distance from zero. This inequality may be interpreted as all values which are less than \(2\) units from zero. On the number line, shown below, all values less than \(2\) units away from zero are indicated.

This graph looks just like the graphs of the three part compound inequalities! When the absolute value is less than a positive number we will determine the solutions to the inequality by changing the problem to a three-part inequality, with the negative value on the "lesser than end" and the positive value on the "greater than end":

as the graph above illustrates.

Now consider

Keep in mind, absolute value is defined as distance from zero. This inequality represents the values which are greater than \(2\) units away from zero. On the number line below we shade all points that are more than \(2\) units away from zero.

When the absolute value is greater than a positive number we find the solution by rewriting the absolute value inequality as two inequalities. The first inequality is used to determine the solutions greater than the positive number, the second inequality yields the solutions which are less than the negative of the number:

as the graph above illustrates.

The solution to an absolute value inequality may also be expressed in interval notation. There is a special symbol to represent the "OR" when two intervals are required. It's called the union: \(\cup.\) We will learn more about unions in Section 4.A.

We can solve absolute value inequalities much like we solved absolute value equations.

Example 2.B.10. Absolute Value Inequality: Less than.

Solve, graph, and give interval notation for the solution.

\begin{align*} \lvert x-2\rvert \color{red}{\lt} 3 \amp\amp\amp \text{Absolute value is less, use three part inequality}\\ -3 \lt x-2\lt 3 \amp\amp\amp \text{Add 2 to all three parts}\\ \underline{+2\phantom{12345}+2\phantom{1}+2} \amp\amp\amp \,\\ -1 \lt x \lt 5 \amp\amp\amp \text{Graph} \end{align*}

Our Graph \(\checkmark\)

Interval notation: \((-1,5)\checkmark\)

Note that the values described in the absolute value inequality \(\lvert x-2\rvert \color{red}{\lt} 3\) are all the values which are within \(3\) units of \(2.\)

Example 2.B.12. Absolute Value Inequality: Greater than or equal.

Solve, graph, and give interval notation for the solution.

\begin{align*} \amp \lvert x+1 \rvert \geq 3 \amp\amp \text{Absolute value is greater than a positive number,}\\ \amp\amp\amp\text{use OR inequality}\\ x+1\leq -3 \amp \text{ OR } x+1\geq 3 \amp\amp \text{Solve both inequalities}\\ \underline{-1\phantom{1}-1}\amp\phantom{123456}\underline{-1\phantom{1}-1} \amp\amp \text{Subtract \(1\) from both sides of each inequality}\\ \color{red}{x\leq -4}\amp\text{ OR }\phantom{1234}\color{blue}{x\geq 2} \amp\amp \text{Graph the solutions} \end{align*}

Our Graph \(\checkmark\)

Interval notation:

\(\left(-\infty,-4\right]\cup\left[2,\infty\right)\checkmark\)

In this case, the absolute value inequality \(\lvert x+1 \rvert \geq 3\) describes all the values which are more than \(3\) units away from \(-1\text{.}\)

In the next example, we will work backwards to find a description for a distance.

Example 2.B.15. Absolute Value Inequality.

Solve, graph, and give interval notation for the solution.

\begin{align*} \amp\lvert 4x-5\rvert \geq 6 \amp\amp \text{Absolute value is greater than a positive number,}\\ \amp\amp\amp\text{use OR inequality}\\ 4x-5\leq -6\amp\text{ OR } 4x-5\geq 6 \amp\amp \text{Solve both inequalities}\\ \underline{+5\phantom{1}+5}\amp\phantom{1234567}\underline{+5\phantom{1}+5} \amp\amp \text{Add \(5\) to both sides of each inequality}\\ 4x\leq -1\amp\text{ OR }\phantom{123} 4x\geq 11 \amp\amp \text{Divide both sides by \(4\)}\\ \overline{4}\phantom{1234}\overline{4}\amp\phantom{1234567}\overline{4}\phantom{12345}\overline{4} \amp\amp \\ \color{blue}{x\leq -\dfrac{1}{4}}\amp\text{ OR }\phantom{1234}\color{red}{x\geq \dfrac{11}{4}} \amp\amp \text{Graph the solutions} \end{align*}

Our Graph \(\checkmark\)

Interval notation:

\(\left(-\infty,-\dfrac{1}{4}\right]\cup\left[\dfrac{11}{4},\infty\right)\checkmark\)

Example 2.B.16. Absolute Value Inequality.

Solve, graph, and give interval notation for the solution.

\begin{align*} \lvert 1-2x\rvert \color{red}{\lt} 3 \amp\amp\amp \text{Absolute value is less,}\\ \amp\amp\amp\text{use three part inequality}\\ -3\lt 1-2x\lt 3 \amp\amp\amp \text{Subtract 1 from all three parts}\\ \underline{-1\phantom{123}-1\phantom{123}-1} \amp\amp\amp \\ -4\lt -2x \lt 2 \amp\amp\amp \text{Divide all three parts by \(-2\)}\\ \overline{-2}\phantom{123}\overline{-2}\phantom{123}\overline{-2} \amp\amp\amp \text{Dividing by a negative means switching the inequality sign}\\ \color{green}{2\gt x \gt -1} \amp\amp\amp \text{Rewrite the inequality (not a necessary step)}\\ \color{green}{-1\lt x \lt 2} \amp\amp\amp \text{Graph} \end{align*}

Our Graph \(\checkmark\)

Interval notation:

\(\left(-1,2\right)\checkmark\)

Recall that there are two special cases when solving absolute value equations. We must consider them for inequalities as well.

Notice in the example above, the absolute value expression is always greater than a negative number. What if the inequality involved a "less than"?

There is one more special case to consider.