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19.2: Solve Compound Inequalities

Last updated

Aug 13, 2020
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- 19.1E: Exercises
- 19.2E: Exercises

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Learning Objectives

By the end of this section, you will be able to:

Solve compound inequalities with “and”
Solve compound inequalities with “or”
Solve applications with compound inequalities

Before you get started, take this readiness quiz.

Simplify: $\frac{2}{5}(x+10)$ .
If you missed this problem, review [link].
Simplify: $−(x−4)$ .
If you missed this problem, review [link].

Solve Compound Inequalities with “and”

Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A compound inequality is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities.

$\begin{array} {lll} {x+3>−4} &{\text{and}} &{4x−5\leq 3} \\ {2(y+1)<0} &{\text{or}} &{y−5\geq −2} \\ \end{array} \nonumber$

COMPOUND INEQUALITY

A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.

To solve a compound inequality with the word “and,” we look for all numbers that make both inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make either inequality true.

Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to both inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets.

$The figure is an illustration of two streets with their intersection shaded$

To find the solution of an "and" compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap.

For the compound inequality $x>−3$ and $x\leq 2$ , we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See Figure $\PageIndex{1}$ .

The figure shows the graph of x is greater than negative 3 with a left parenthesis at negative 3 and shading to its right, the graph of x is less than or equal to 2 with a bracket at 2 and shading to its left, and the graph of x is greater than negative 3 and x is less than or equal to 2 with a left parenthesis at negative 3 and a right parenthesis at 2 and shading between negative 3 and 2. Negative 3 and 2 are marked by lines on each number line. — Figure $\PageIndex{1}$

We can see that the numbers between $−3$ and $2$ are shaded on both of the first two graphs. They will then be shaded on the solution graph.

The number $−3$ is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph.

The number two is shaded on both the first and second graphs. Therefore, it is be shaded on the solution graph.

This is how we will show our solution in the next examples.

Example $\PageIndex{1}$

Solve $6x−3<9$ and $2x+7\geq 3$ . Graph the solution and write the solution in interval notation.

Solution

	$6x−3<9$	and	$2x+9\geq 3$
Step 1. Solve each inequality.	$6x−3<9$		$2x+9\geq 3$
	$6x<12$		$2x\geq −6$
	$x<2$	and	$x\geq −3$
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.	$.$
Step 3. Write the solution in interval notation.	$[−3,2)$
All the numbers that make both inequalities true are the solution to the compound inequality.

Try It $\PageIndex{2}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $4x−7<9$ and $5x+8\geq 3$ .

Answer: $The solution is negative 1 is less than or equal to x which is less than 4. On a number line it is shown with a closed circle at negative 1 and an open circle at 4 with shading in between the closed and open circles. Its interval notation is negative 1 to 4 within a bracket and a parenthesis.$

Try It $\PageIndex{3}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $3x−4<5$ and $4x+9\geq 1$ .

Answer: $The solution is negative 2 is less than or equal to x which is less than 3. On a number line it is shown with a closed circle at negative 2 and an open circle at 3 with shading in between the closed and open circles. Its interval notation is negative 2 to 3 within a bracket and a parenthesis.$

SOLVE A COMPOUND INEQUALITY WITH “AND.”

Solve each inequality.
Graph each solution. Then graph the numbers that make both inequalities true.
This graph shows the solution to the compound inequality.
Write the solution in interval notation.

Example $\PageIndex{4}$

Solve $3(2x+5)\leq 18$ and $2(x−7)<−6$ . Graph the solution and write the solution in interval notation.

Solution

	$3(2x+5)\leq 18$	and	$2(x−7)<−6$
Solve each inequality.	$6x+15\leq 18$		$2x−14<−6$
	$6x\leq 3$		$2x<8$
	$x\leq \frac{1}{2}$	and	$x<4$
Graph each solution.	$.$
Graph the numbers that make both inequalities true.	$.$
Write the solution in interval notation.	$(−\infty, \frac{1}{2}]$

Try It $\PageIndex{5}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $2(3x+1)\leq 20$ and $4(x−1)<2$ .

Answer: $The solution is x is less than three-halves. On a number line it is shown with an open circle at three-halves with shading to its left. Its interval notation is negative infinity to three-halves within a parentheses.$

Try It $\PageIndex{6}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $5(3x−1)\leq 10$ and $4(x+3)<8$ .

Answer: $The solution is x is less than negative 1. On a number line it is shown with an open circle at 1 with shading to its left. Its interval notation is negative infinity to negative 1 within parentheses.$

Example $\PageIndex{7}$

Solve $\frac{1}{3}x−4\geq −2$ and $−2(x−3)\geq 4$ . Graph the solution and write the solution in interval notation.

Solution

	$\frac{1}{3}x−4\geq −2$	and	$−2(x−3)\geq 4$
Solve each inequality.	$\frac{1}{3}x−4\geq −2$		$−2x+6\geq 4$
	$\frac{1}{3}x\geq 2$		$−2x\geq −2$
	$x\geq 6$	and	$x\leq 1$
Graph each solution.	$.$
Graph the numbers that make both inequalities true.	$.$
	This is a contradiction. There are no numbers that make both inequalities true, so there is no solution.

Try It $\PageIndex{8}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{1}{4}x−3\geq −1$ and $−3(x−2)\geq 2$ .

Answer: $The inequality is a contradiction. So, there is no solution. As a result, there is no graph of the number line or interval notation.$

Try It $\PageIndex{9}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{1}{5}x−5\geq −3$ and $−4(x−1)\geq −2$ .

Answer: $The inequality is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.$

Sometimes we have a compound inequality that can be written more concisely. For example, $a<x$ and $x<b$ can be written simply as $a<x<b$ and then we call it a double inequality. The two forms are equivalent.

DOUBLE INEQUALITY

A double inequality is a compound inequality such as $a<x<b$ . It is equivalent to $a<x$ and $x<b$ .

$\text{Other forms:} \quad \begin{array} {lllll} {a<x<b} &{\text{is equivalent to }} &{a<x} &{\text{and}} &{x<b} \\ {a\leq x\leq b} &{\text{is equivalent to }} &{a\leq x} &{\text{and}} &{x\leq b} \\ {a>x>b} &{\text{is equivalent to }} &{a>x} &{\text{and}} &{x>b} \\ {a\geq x\geq b} &{\text{is equivalent to }} &{a\geq x} &{\text{and}} &{x\geq b} \\ \end{array} \nonumber$

To solve a double inequality we perform the same operation on all three “parts” of the double inequality with the goal of isolating the variable in the center.

Example $\PageIndex{10}$

Solve $−4\leq 3x−7<8$ . Graph the solution and write the solution in interval notation.

Solution

	$-4 \leq 3x - 7 < 8$
Add 7 to all three parts.	$-4 \,{\color{red}{+\, 7}} \leq 3x - 7 \,{\color{red}{+ \,7}} < 8 \,{\color{red}{+ \,7}}$
Simplify.	$3 \le 3x < 15$
Divide each part by three.	$\dfrac{3}{\color{red}{3}} \leq \dfrac{3x}{\color{red}{3}} < \dfrac{15}{\color{red}{3}}$
Simplify.	$1 \leq x < 5$
Graph the solution.	$.$
Write the solution in interval notation.	$[1, 5)$

When written as a double inequality, $1\leq x<5$ , it is easy to see that the solutions are the numbers caught between one and five, including one, but not five. We can then graph the solution immediately as we did above.

Another way to graph the solution of $1\leq x<5$ is to graph both the solution of $x\geq 1$ and the solution of $x<5$ . We would then find the numbers that make both inequalities true as we did in previous examples.

Try It $\PageIndex{11}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $−5\leq 4x−1<7$ .

Answer: $The solution is negative 1 is less than or equal to x which is less than 2. Its graph has a closed circle at negative 1 and an open circle at 2 with shading between the closed and open circles. Its interval notation is negative 1 to 2 within a bracket and a parenthesis.$

Try It $\PageIndex{12}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $−3<2x−5\leq 1$ .

Answer: $The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 with shading between the closed and open circles. Its interval notation is negative 1 to 3 within a parenthesis and a bracket.$

Solve Compound Inequalities with “or”

To solve a compound inequality with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make either inequality true.

Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.

To write the solution in interval notation, we will often use the union symbol, $\cup$ , to show the union of the solutions shown in the graphs.

SOLVE A COMPOUND INEQUALITY WITH “OR.”

Solve each inequality.
Graph each solution. Then graph the numbers that make either inequality true.
Write the solution in interval notation.

Example $\PageIndex{13}$

Solve $5−3x\leq −1$ or $8+2x\leq 5$ . Graph the solution and write the solution in interval notation.

Solution

	$5−3x\leq −1$	or	$8+2x\leq 5$
Solve each inequality.	$5−3x\leq −1$		$8+2x\leq 5$
	$−3x\leq −6$		$2x\leq −3$
	$x\geq 2$	or	$x\leq −\frac{3}{2}$
Graph each solution.	$.$
Graph numbers that make either inequality true.	$.$
	$(−\infty,−\frac{3}{2}]\cup[2,\infty)$

Try It $\PageIndex{14}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $1−2x\leq −3$ or $7+3x\leq 4$ .

Answer: $The solution is x is greater than or equal to 2 or x is less than or equal to 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 2 and infinity within a bracket and a parenthesis.$

Try It $\PageIndex{15}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $2−5x\leq −3$ or $5+2x\leq 3$ .

Answer: $The solution is x is greater than or equal to 1 or x is less than or equal to negative 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 1 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 1 and infinity within a bracket and a parenthesis.$

Example $\PageIndex{16}$

Solve $\frac{2}{3}x−4\leq 3$ or $\frac{1}{4}(x+8)\geq −1$ . Graph the solution and write the solution in interval notation.

Solution

	$\frac{2}{3}x−4\leq 3$	or	$\frac{1}{4}(x+8)\geq −1$
Solve each inequality.	$3(\frac{2}{3}x−4)\leq 3(3)$		$4⋅\frac{1}{4}(x+8)\geq 4⋅(−1)$
	$2x−12\leq 9$		$x+8\geq −4$
	$2x\leq 21$		$x\geq −12$
	$x\leq \frac{21}{2}$
	$x\leq \frac{21}{2}$	or	$x\geq −12$
Graph each solution.	$.$
Graph numbers that make either inequality true.	$.$
	The solution covers all real numbers.
	$(−\infty ,\infty )$

Try It $\PageIndex{17}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{5}x−7\leq −1$ or $\frac{1}{3}(x+6)\geq −2$ .

Answer: $The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.$

Try It $\PageIndex{18}$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{4}x−3\leq 3$ or $\frac{2}{5}(x+10)\geq 0$ .

Answer: $The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.$

Solve Applications with Compound Inequalities

Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications.

Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a compound inequality. Last, we will solve the compound inequality.

Example $\PageIndex{19}$

Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $57.06 and $171.02. How many hcf can the owner use if he wants his usage to stay in the normal range?

Solution

Identify what we are looking for.	The number of hcf he can use and stay in the “normal usage” billing range.
Name what we are looking for.	Let x= the number of hcf he can use.
Translate to an inequality.	Bill is $24.72 plus $1.54 times the number of hcf he uses or $24.72+1.54x$ .
	$\color{Cerulean}{\underbrace{\color{black}{\text{His bill will be between or equal to }$57.06\text{ and }$171.02.}}}$ $57.06 \leq 24.72 + 1.54x \leq 171.02$
Solve the inequality.	$57.06 \leq 24.72 + 1.54x \leq 171.02$ $57.06 \,{\color{red}{- \,24.72}}\leq 24.72 \,{\color{red}{- \,24.72}} + 1.54x \leq 171.02 \,{\color{red}{- \,24.72}}$ $32.34 \leq 1.54x \leq 146.3$ $\dfrac{32.34}{\color{red}{1.54}} \leq \dfrac{1.54x}{\color{red}{1.54}} \leq \dfrac{146.3}{\color{red}{1.54}}$ $21 \leq x \leq 95$
Answer the question.	The property owner can use $21–95$ hcf and still fall within the “normal usage” billing range.

Try It $\PageIndex{20}$

Due to the drought in California, many communities now have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to $31.32 and $51.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?

Answer: The homeowner can use $5–20$ hcf and still fall within the “conservation usage” billing range.

Try It $\PageIndex{21}$

During the winter, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $49.36 and $86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range?

Answer: The homeowner can use $16–40$ hcf and still fall within the “normal usage” billing range.

Access this online resource for additional instruction and practice with solving compound inequalities.

Compound inequalities

Key Concepts

How to solve a compound inequality with “and”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make both inequalities true. This graph shows the solution to the compound inequality.
3. Write the solution in interval notation.
Double Inequality
- A double inequality is a compound inequality such as $a<x<b$ . It is equivalent to $a<x$ and $x<b.$
  
  Other forms: $\begin{align*} a<x<b & & \text{is equivalent to} & & a<x\;\text{and}\;x<b \\ a≤x≤b & & \text{is equivalent to} & & a≤x\;\text{and}\;x≤b \\ a>x>b & & \text{is equivalent to} & & a>x\;\text{and}\;x>b \\ a≥x≥b & & \text{is equivalent to} & & a≥x\;\text{and}\;x≥b \end{align*}$
How to solve a compound inequality with “or”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make either inequality true.
3. Write the solution in interval notation.

Glossary

compound inequality: A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

Learning Objectives

Solve Compound Inequalities with “and”

COMPOUND INEQUALITY

Example 19.2.1\PageIndex{1}

Try It 19.2.2\PageIndex{2}

Try It 19.2.3\PageIndex{3}

SOLVE A COMPOUND INEQUALITY WITH “AND.”

Example 19.2.4\PageIndex{4}

Try It 19.2.5\PageIndex{5}

Try It 19.2.6\PageIndex{6}

Example 19.2.7\PageIndex{7}

Try It 19.2.8\PageIndex{8}

Try It 19.2.9\PageIndex{9}

DOUBLE INEQUALITY

Example 19.2.10\PageIndex{10}

Try It 19.2.11\PageIndex{11}

Try It 19.2.12\PageIndex{12}

Solve Compound Inequalities with “or”

SOLVE A COMPOUND INEQUALITY WITH “OR.”

Example 19.2.13\PageIndex{13}

Try It 19.2.14\PageIndex{14}

Try It 19.2.15\PageIndex{15}

Example 19.2.16\PageIndex{16}

Try It 19.2.17\PageIndex{17}

Try It 19.2.18\PageIndex{18}

Solve Applications with Compound Inequalities

Example 19.2.19\PageIndex{19}

Try It 19.2.20\PageIndex{20}

Try It 19.2.21\PageIndex{21}

Key Concepts

Glossary

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