6.2.3 Solving Compound Inequalities

Learning Outcomes

  • Solve compound inequalities with OR (Unions) – express solutions graphically, in set-builder notation and with interval notation
  • Solve compound inequalities with AND (Intersections) – express solutions graphically, in set-builder notation and with interval notation

Key words

  • Compound Inequality: two inequalities joined as a union or an intersection
  • Union: values exist in either set. Key word is “or”.
  • Intersection: values exist in both sets. Key word is “and”.

Compound Inequalities

A compound inequality is two inequalities joined together with the word “or” or the word “and”. When the two inequalities are joined with “or”, the solution set must satisfy either equation. For example, [latex]x\lt 2 \text{ or } x\ge 5[/latex] tells us that the [latex]x[/latex]-values must be either less than [latex]2[/latex] or greater than or equal to [latex]5[/latex]. On the other hand, [latex]x\ge 2\text{ and }x\lt 5[/latex] tells us that the [latex]x[/latex]-values must be greater than or equal to [latex]2[/latex] and also less than [latex]5[/latex]. When “or” is used, the solution set is a union of the solution sets of both inequalities. When “and” is used, the solution set is an intersection of the solution sets of both inequalities.

Unions

The solution of a compound inequality that consists of two inequalities joined with the word or is the union of the solutions of each inequality. Unions allow us to create a new set from two that may or may not have elements in common.

The following example shows how to solve a one-step inequality in the or form. Note how each inequality is treated independently until the end, where the solution is described in terms of both inequalities. We will use the same properties to solve compound inequalities that we used to solve single inequalities.

Example

Solve for [latex]x[/latex].  [latex]3x–1<8[/latex] or [latex]x–5>0[/latex]

Solution:

Solve each inequality by isolating the variable.

[latex]\displaystyle \begin{array}{r}x-5>0\,\,\,\,\,\,\,\,\textit{or}\,\,\,\,\,\,\,\,\,\,3x-1<8\,\,\\\underline{\,\,\,+5\,\,+5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,+1\,\,+1}\\x\,\,>5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{3x}\,\,\,<\underline{9}\\{3}\,\,\,\,\,\,\,\,\,\,\,\,\,{3}\\x<3\,\,\,\\x>5\,\,\,\,\textit{or}\,\,\,\,x<3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Set-builder notation: [latex]\{\;x\large\;|\;\normalsize x\lt 3\text{ or }x\gt 5,\;y\in\mathbb{R}\}[/latex]

Interval notation: [latex]\left(-\infty, 3\right)\cup\left(5,\infty\right)[/latex]

The solution to this compound inequality can also be shown graphically. Sometimes it helps to draw the graph first, before writing the solution using interval notation.

Number line. Open red circle on 3 and red highlight through all numbers less than 3. Open blue circle on 5 and blue highlight on all numbers greater than 5.

 

Example

Solve for [latex]y[/latex].  [latex]2y+7\lt13[/latex] or [latex]−3y–2\lt10[/latex]

Solution:

Solve each inequality separately.

[latex]\displaystyle \begin{array}{r}2y+7<13\,\,\,\,\,\,\,\,\textit{or}\,\,\,\,\,\,\,\,\,\,-3y-2\lt 10\\\underline{\,\,\,-7\,\,-7}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,+2\,\,\,+2}\\\underline{2y}<\underline{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3y}<\underline{12}\\{2}\,\,\,\,\,\,\,\,\,\,\,{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{-3}\,\,\,\,\,\,\,\,\,\,\,{-3}\\y<3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\gt -4\\y<3\,\,\,\,\textit{or}\,\,\,\,y\gt -4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

The inequality sign is reversed with division by a negative number.

Since [latex]y[/latex] could be less than [latex]3[/latex] or greater than [latex]−4,\;y[/latex] could be any real number. Graphing the inequality helps with this interpretation.

Graph:

Line graph

The individual graphs show empty dots at [latex]y=3[/latex] and [latex]y=-4[/latex], but these graphs combine to all real numbers.

Set-builder notation: [latex]\{\;y\large\;|\;\normalsize \;y\in\mathbb{R}\}[/latex]

Interval notation: [latex]x\in\left(-\infty,\infty\right)[/latex]

 

In the last example, the final answer included solutions whose intervals overlapped. This caused the answer to include all numbers on the number line. In words, we call this solution “all real numbers”. Any real number will produce a true statement for either [latex]y<3\text{ or }y\gt -4[/latex] when it is substituted for y.

Example

Solve for [latex]z[/latex].

[latex]5z–3\gt−18[/latex] or [latex]−2z–1\gt15[/latex]

 

Solution:

Solve each inequality separately. Combine the solutions.

[latex]\displaystyle \begin{array}{r}5z-3>18\,\,\,\,\,\,\,\,\textit{or}\,\,\,\,\,\,\,\,\,\,-2z-1>15\\\underline{\,\,\,+3\,\,\,+3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,+1\,\,\,+1}\\\underline{5z}>\underline{-15}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-2z}>\underline{16}\\{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{-2}\,\,\,\,\,\,\,\,\,\,\,{-2}\\z>-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z<-8\\z>-3\,\,\,\,\textit{or}\,\,\,\,z<-8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Set-builder notation: [latex]\{\;z\large\;|\normalsize z>-3\,\,\text{or}\,\,z<-8,\;z\in\mathbb{R}\}[/latex] Interval notation: [latex]x\in\left(-\infty,-8\right)\cup\left(-3,\infty\right)[/latex] Note how we write the intervals with the one containing the most negative solutions first, then move to the right on the number line. [latex]z<-8[/latex] has solutions that continue all the way to the left on the number line, whereas [latex]x>-3[/latex] has solutions that continue all the way to the right.

Graph:Number line. Red open circle on negative 8 and red highlight on all numbers less than negative 8. Open blue circle on negative 3 and blue highlight through all numbers greater than negative 3.

 

The following video contains an example of solving a compound inequality involving or and drawing the associated graph.

Try It

Intersections

The solution of a compound inequality that consists of two inequalities joined with the word and is the intersection of the solutions of each inequality. In other words, both statements must be true at the same time. The solution to an and compound inequality are all the solutions that the two inequalities have in common. This is where the two graphs overlap.

In this section we will see more examples where we have to simplify the compound inequalities before we can express their solutions graphically or with an interval.

Example

Solve for x. [latex]\displaystyle 1-4x\le 21\,\,\,\,\text{and}\,\,\,\,5x+2\ge22[/latex]

 

Solution:

Solve each inequality for x. Determine the intersection of the solutions.

[latex]\displaystyle \begin{array}{r}\,\,\,1-4x\le 21\,\,\,\,\,\,\,\,\text{AND}\,\,\,\,\,\,\,5x+2\ge 22\\\underline{-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,-2\,\,\,\,-2}\\\,\,\,\,\,\underline{-4x}\leq \underline{20}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{5x}\,\,\,\,\,\,\,\ge \underline{20}\\\,\,\,\,\,{-4}\,\,\,\,\,\,\,{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{5}\,\,\\\,\,\,\,\,\,\,\,\,\,\,x\ge -5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\ge 4\,\,\,\,\\\\x\ge -5\,\text{and}\,\,x\ge 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

The number line below shows the graphs of the two inequalities in the problem. The solution to the compound inequality is [latex]x\geq4[/latex], since this is where the two graphs overlap.

Number line. Closed blue circle on negative 5 and blue arrow through all numbers greater than negative 5. This blue arrow is labeled x is greater than or equal to negative 5. Closed red circle on 4 and red arrow through all numbers greater than 4. This red line is labeled x is greater than or equal to 4.

Answer

Set-builder notation: [latex]\{x\;\large |\;\normalsize x\ge 4,\;x\in\mathbb{R}\}[/latex]

Interval notation: [latex]x\in\left[4,\infty\right)[/latex]

Graph:

Number line. Closed purple circle (overlapping red and blue circles) on 4 and purple arrow through all numbers greater than 4. Purple line is labeled x is greater than or equal to 4.

EXample

Solve for x:  [latex]\displaystyle {5}{x}-{2}\le{3}\text{ and }{4}{x}{+7}>{3}[/latex]

Solution:

Solve each inequality separately. Find the overlap between the solutions.

[latex]\displaystyle \begin{array}{l}\,\,\,5x-2\le 3\,\,\,\,\,\,\,\,\,\text{AND}\,\,\,\,\,\,\,4x+7>\,\,\,\,3\\\underline{\,\,\,\,\,\,\,\,\,\,\,+2\,\,+2\,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,\,\,\,\,-7\,\,\,\,\,\,-7}\\\,\,\frac{5x}{5}\,\,\,\,\,\,\,\,\le \frac{5}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4x}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,>\frac{-4}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x>-1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 1\,\,\,\,\text{and}\,\,\,\,x>-1\end{array}[/latex]

Answer

Set-builder notation: [latex]\{\;x\large\; |\normalsize\;-1\le{x}\le{1},\;x\in\mathbb{R}\}[/latex]

Interval notation: [latex]x\in\left(-1,1\right][/latex]

Graph:image

 

Notice that the two inequalities [latex]x\gt -1[/latex] and [latex]x\leq 1[/latex] can be combined and written as [latex]-1\lt x\leq 1[/latex]. This is best seen on the graph where the solution set lies between [latex]-1[/latex] and [latex]1[/latex]. In other words, [latex]x[/latex] is trapped between [latex]-1[/latex] and [latex]1[/latex]. All intersection inequalities [latex]x\gt a[/latex] and [latex]x\lt b[/latex] can be written in the form [latex]a\lt x\lt b[/latex] provided [latex]a\lt b[/latex]. If [latex]a\gt b[/latex] there is no graphical overlap which means there is no solution.

Try It

Intersections in the form [latex]a

A compound inequality that is an intersection can be written in the form [latex]a\lt x\lt b[/latex] rather than [latex]x\gt a \text{ and }x\lt b[/latex] provided [latex]a\lt b[/latex]. If [latex]a\gt b[/latex] there is no solution. Rather than splitting a compound inequality in the form of  [latex]a and [latex]x\lt b[/latex], we can more quickly solve the inequality by applying the properties of inequality to all three segments of the compound inequality.

Example

Solve for x. [latex]3\lt2x+3\leq 7[/latex]

 

Solution:

Isolate the variable by subtracting [latex]3[/latex] from all [latex]3[/latex] parts of the inequality, then dividing each part by  [latex]2[/latex].

[latex]\begin{array}{r}\,\,\,\,3\,\,\lt\,\,2x+3\,\,\leq \,\,\,\,7\\\underline{\,-3}\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,-3}\,\,\,\,\,\,\,\,\underline{\,-3}\,\\\,\,\,\,\,\underline{\,0\,}\,\,\lt\,\,\,\,\underline{2x}\,\,\,\,\,\,\,\,\leq\,\,\,\underline{\,4\,}\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\\\,\,\,\,\,\,\,\,\,\,0\lt x\leq 2\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Answer

Set-builder notation: [latex]\{\;x\;\large |\normalsize\; 0\lt{x}\le 2,\;x\in\mathbb{R}\}[/latex]

Intervalotation n: [latex]x\in\left(0,2\right][/latex]

Graph:

Open dot on zero, closed dot on 2, and line through all numbers between zero and two.

 

The video below shows another example of how to solve an inequality in the form [latex]a

 

The solution to a compound inequality with and is always the overlap between the solution to each inequality. There are three possible outcomes for compound inequalities joined by the word and:

Case 1:
Description The solution could be all the values between two endpoints
Inequalities [latex]x\le{1}[/latex] and [latex]x\gt{-1}[/latex], or as a bounded inequality: [latex]{-1}\lt{x}\le{1}[/latex]
Interval [latex]\left(-1,1\right][/latex]
Graphs Number line. Open blue circle on negative 1 and blue arrow through all numbers greater than negative 1. The blue arrow represents x is greater than negative 1. Closed red circle on 1 and red arrow through all numbers less than 1. Red arrow written x is less than or equal to 1.

Number line. Open blue circle on negative 1. Closed red circle on 1. Overlapping red and blue lines between negative 1 and 1 that represents negative 1 is less than x is less than or equal to 1.
Case 2:
Description The solution could begin at a point on the number line and extend in one direction.
Inequalities [latex]x\gt3[/latex] and [latex]x\ge4[/latex]
Interval [latex]\left[4,\infty\right)[/latex]
Graphs  Number line. Blue open circle on negative 3 and blue arrow through all numbers greater than negative 3. Blue arrow represents x is greater than negative three. Closed red circle on 4 and red arrow through all numbers greater than 4. The red arrow respresents x is greater than or equal to 4.

Number line. Closed circle on 4 and arrow through all numbers greater than 4. The arrow represents x is greater than or equal to 4.
Case 3:
 Description In cases where there is no overlap between the two inequalities, there is no solution to the compound inequality
 Inequalities [latex]x\lt{-3}[/latex] and [latex]x\gt{3}[/latex]
 Intervals [latex]\left(-\infty,-3\right)[/latex] and [latex]\left(3,\infty\right)[/latex]
 Graph Number line. Open red circle on negative 3 and red arrow through all numbers less than negative 3. Red arrow represents x is less than negative 3. Open blue circle on 3 and blue arrow through all numbers greater than 3. Blue arrow represents x is greater than 3.

In the example below, there is no solution to the compound inequality because there is no overlap between the inequalities.

Example

Solve for x. [latex]x+2>5[/latex] and [latex]x+4<5[/latex] Solution:

Solve each inequality separately.

[latex]\displaystyle \begin{array}{l}x+2>5\,\,\,\,\,\,\,\,\,\text{AND}\,\,\,\,\,\,\,x+4<5\,\,\,\,\\\underline{\,\,\,\,\,-2\,-2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,-4\,-4}\\x\,\,\,\,\,\,\,\,>\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,<\,1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,x>3\,\,\,\,\text{and}\,\,\,\,x<1\end{array}[/latex]

Find the overlap between the solutions.

Number line. Red open circle is on 1 and red arrow through all numbers less than 1. Red arrow is labeled x is less than 1. Open blue circle on 3 and blue arrow through all numbers greater than 3. Blue arrow represents x is greater than 3.

Answer

There is no overlap between [latex]\displaystyle x>3[/latex] and [latex]x<1[/latex], so there is no solution.

Summary

A compound inequality is a statement of two inequality statements linked together either by the word or or by the word and. Sometimes, an and compound inequality is shown symbolically, like [latex]aand. Because compound inequalities represent either a union or intersection of the individual inequalities, graphing them on a number line can be a helpful way to see or check a solution. Compound inequalities can be manipulated and solved in much the same way any inequality is solved, by paying attention to the properties of inequalities and the rules for solving them.