Using Interval Notation

Indicating the solution to an inequality such as[latex]\,x\ge 4\,[/latex]can be achieved in several ways.

We can use a number line as shown in (Figure). The blue ray begins at[latex]\,x=4\,[/latex]and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.

Figure 2.

We can use set-builder notation:[latex]\,\left\{x|x\ge 4\right\},[/latex]which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.

The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to[latex]\,x\ge 4\,[/latex]are represented as[latex]\,\left[4,\infty \right).\,[/latex]This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are[latex]\,\left[-2,6\right),[/latex]or all numbers between[latex]\,-2\,[/latex]and[latex]\,6,[/latex]including[latex]\,-2,[/latex]but not including[latex]\,6;[/latex][latex]\left(-1,0\right),[/latex]all real numbers between, but not including[latex]\,-1\,[/latex]and[latex]\,0;[/latex]and[latex]\,\left(-\infty ,1\right],[/latex]all real numbers less than and including[latex]\,1.\,[/latex](Figure) outlines the possibilities.

Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

Understanding Compound Inequalities

A compound inequality includes two inequalities in one statement. A statement such as[latex]\,4<x\le 6\,[/latex]means[latex]\,4<x\,[/latex]and[latex]\,x\le 6.\,[/latex]There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts:[latex]\,3+x>7x-2>5x-10.[/latex]

Show Solution

Let’s try the first method. Write two inequalities:

[latex]\begin{array}{lll}3+x>7x-2\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & 7x-2>5x-10\hfill \\ \phantom{\rule{1.2em}{0ex}}3>6x-2\hfill & \hfill & 2x-2>-10\hfill \\ \phantom{\rule{1.2em}{0ex}}5>6x\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}2x>-8\hfill \\ \phantom{\rule{1.2em}{0ex}}\frac{5}{6}>x\hfill & \hfill & \phantom{\rule{2.5em}{0ex}}x>-4\hfill \\ \phantom{\rule{1.3em}{0ex}}x<\frac{5}{6}\hfill & \hfill & \phantom{\rule{1.5em}{0ex}}-4<x\hfill \end{array}[/latex]

The solution set is[latex]\,-4<x<\frac{5}{6}\,[/latex]or in interval notation[latex]\,\left(-4,\frac{5}{6}\right).\,[/latex]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. See (Figure).

Figure 3.

Solving Absolute Value Inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at[latex]\,\left(-x,0\right)\,[/latex]has an absolute value of[latex]\,x,[/latex]as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all [latex]\,x[/latex]–values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between[latex]\,x\,[/latex]and 600 is less than 200. We represent the distance between[latex]\,x\,[/latex]and 600 as[latex]\,|x-600|,[/latex]and therefore,[latex]\,|x-600|\le 200\,[/latex]or

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Determining a Number within a Prescribed Distance

Describe all values[latex]\,x\,[/latex]within a distance of 4 from the number 5.

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We want the distance between[latex]\,x\,[/latex]and 5 to be less than or equal to 4. We can draw a number line, such as in (Figure), to represent the condition to be satisfied.

Figure 4.

The distance from[latex]\,x\,[/latex]to 5 can be represented using an absolute value symbol,[latex]\,|x-5|.\,[/latex]Write the values of[latex]\,x\,[/latex]that satisfy the condition as an absolute value inequality.

[latex]|x-5|\le 4[/latex]

We need to write two inequalities as there are always two solutions to an absolute value equation.

[latex]\begin{array}{lll}x-5\le 4\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & x-5\ge -4\hfill \\ \phantom{\rule{1.8em}{0ex}}x\le 9\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}x\ge 1\hfill \end{array}[/latex]

If the solution set is[latex]\,x\le 9\,[/latex]and[latex]\,x\ge 1,[/latex]then the solution set is an interval including all real numbers between and including 1 and 9.

So[latex]\,|x-5|\le 4\,[/latex]is equivalent to[latex]\,\left[1,9\right]\,[/latex]in interval notation.

Using a Graphical Approach to Solve Absolute Value Inequalities

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

Show Solution

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

[latex]\begin{array}{ll}-\frac{1}{2}|4x-5|<-3\hfill & \phantom{\rule{2em}{0ex}}\text{Multiply both sides by –2, and reverse the inequality}.\hfill \\ \phantom{\rule{1.5em}{0ex}}|4x-5|>6\hfill & \hfill \end{array}[/latex]

Next, we solve for the equality [latex]|4x-5|=6.[/latex]

[latex]\begin{array}{lll}4x-5=6\hfill & \hfill & 4x-5=-6\hfill \\ \phantom{\rule{1.9em}{0ex}}4x=11\hfill & \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\hfill & \phantom{\rule{1.9em}{0ex}}4x=-1\hfill \\ \phantom{\rule{2em}{0ex}}x=\frac{11}{4}\hfill & \hfill & \phantom{\rule{2em}{0ex}}x=-\frac{1}{4}\hfill \end{array}[/latex]

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[latex]\,x=-\frac{1}{4}\,[/latex]and[latex]\,x=\frac{11}{4},[/latex]and that the graph opens downward. See (Figure).

Figure 5.