Solve Compound Inequalities

Lynn Marecek

82 Solve Compound Inequalities

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: $\frac{2}{5}\phantom{\rule{0.2em}{0ex}}\left(x+10\right).$

If you missed this problem, review (Figure).
Simplify: $\text{−}\left(x-4\right).$

If you missed this problem, review (Figure).

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.

$x+3>-4\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}4x-5\le 3$

$2\left(y+1\right)<0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y-5\ge -2$

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 how 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 the 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 $x\le 2,$ 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).

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.

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.

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

	$6x-3<9\phantom{\rule{0.55em}{0ex}}$	and	$2x+9\ge 3\phantom{\rule{0.65em}{0ex}}$
Step 1. Solve each inequality.	$6x-3<9\phantom{\rule{0.55em}{0ex}}$		$2x+9\ge 3\phantom{\rule{0.65em}{0ex}}$
	$6x<12$		$2x\ge -6$
	$x<2\phantom{\rule{0.55em}{0ex}}$	and	$x\ge -3$
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.	$\left[-3,2\right)$
All the numbers that make both inequalities true are the solution to the compound inequality.

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

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.

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

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.

Solve $3\left(2x+5\right)\le 18$ and $2\left(x-7\right)<-6.$ Graph the solution and write the solution in interval notation.

	$3\left(2x+5\right)\le 18$	and	$2\left(x-7\right)<-6$
Solve each inequality.	$6x+15\le 18$		$2x-14<-6$
	$6x\le 3\phantom{\rule{0.5em}{0ex}}$		$2x<8\phantom{\rule{0.65em}{0ex}}$
	$x\le \frac{1}{2}\phantom{\rule{0.4em}{0ex}}$	and	$x<4\phantom{\rule{0.65em}{0ex}}$
Graph each solution.
Graph the numbers that make both inequalities true.
Write the solution in interval notation.	$\left(\text{−}\infty ,\frac{1}{2}\right]$

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

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.

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

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.

Solve $\frac{1}{3}x-4\ge -2$ and $-2\left(x-3\right)\ge 4.$ Graph the solution and write the solution in interval notation.

	$\frac{1}{3}x-4\ge -2$	and	$-2\left(x-3\right)\ge 4\phantom{\rule{0.75em}{0ex}}$
Solve each inequality.	$\frac{1}{3}x-4\ge -2$		$-2x+6\ge 4\phantom{\rule{0.75em}{0ex}}$
	$\frac{1}{3}x\ge 2\phantom{\rule{0.65em}{0ex}}$		$-2x\ge -2$
	$x\ge 6\phantom{\rule{0.65em}{0ex}}$	and	$x\le 1\phantom{\rule{0.75em}{0ex}}$
Graph each solution.
Graph the numbers that make both inequalities true.
	$\phantom{\rule{8em}{0ex}}$ There are no numbers that make both inequalities true. $\phantom{\rule{8em}{0ex}}$ This is a contradiction so there is no solution.

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

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

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

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 $x<b.$

$\begin{array}{cccc}\text{Other forms:}\hfill & & & \begin{array}{ccccccccccccc}a<x<b\hfill & & & \text{is equivalent to}\hfill & & & a<x\hfill & & & \text{and}\hfill & & & x<b\hfill \\ a\le x\le b\hfill & & & \text{is equivalent to}\hfill & & & a\le x\hfill & & & \text{and}\hfill & & & x\le b\hfill \\ a>x>b\hfill & & & \text{is equivalent to}\hfill & & & a>x\hfill & & & \text{and}\hfill & & & x>b\hfill \\ a\ge x\ge b\hfill & & & \text{is equivalent to}\hfill & & & a\ge x\hfill & & & \text{and}\hfill & & & x\ge b\hfill \end{array}\hfill \end{array}$

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.

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


Add 7 to all three parts.
Simplify.
Divide each part by three.
Simplify.
Graph the solution.
Write the solution in interval notation.

When written as a double inequality, $1\le x<5,$ 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 $1\le x<5$ is to graph both the solution of $x\ge 1$ and the solution of $x<5.$ We would then find the numbers that make both inequalities true as we did in previous examples.

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

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.

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

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.

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

	$5-3x\le -1$	or	$8+2x\le 5\phantom{\rule{1.25em}{0ex}}$
Solve each inequality.	$5-3x\le -1$		$8+2x\le 5\phantom{\rule{1.25em}{0ex}}$
	$-3x\le -6$		$\phantom{\rule{3em}{0ex}}2x\le -3\phantom{\rule{0.1em}{0ex}}$
	$x\ge 2\phantom{\rule{0.65em}{0ex}}$	or	$\phantom{\rule{3.2em}{0ex}}x\le -\frac{3}{2}$
Graph each solution.
Graph numbers that make either inequality true.
	$\left(\text{−}\infty ,-\frac{3}{2}\right]\cup \left[2,\infty \right)$

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

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.

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

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.

Solve $\frac{2}{3}x-4\le 3$ or $\frac{1}{4}\left(x+8\right)\ge -1.$ Graph the solution and write the solution in interval notation.

	$\frac{2}{3}x-4\le 3\phantom{\rule{1.2em}{0ex}}$	or	$\frac{1}{4}\left(x+8\right)\ge -1\phantom{\rule{1.8em}{0ex}}$
Solve each inequality.	$\phantom{\rule{0.4em}{0ex}}3\left(\frac{2}{3}x-4\right)\le 3\left(3\right)$		$4·\frac{1}{4}\left(x+8\right)\ge 4·\left(-1\right)$
	$2x-12\le 9\phantom{\rule{1.2em}{0ex}}$		$x+8\ge -4\phantom{\rule{1.75em}{0ex}}$
	$2x\le 21\phantom{\rule{0.7em}{0ex}}$		$x\ge -12\phantom{\rule{1.25em}{0ex}}$
	$x\le \frac{21}{2}\phantom{\rule{0.55em}{0ex}}$
	$x\le \frac{21}{2}\phantom{\rule{0.55em}{0ex}}$	or	$x\ge -12\phantom{\rule{1.25em}{0ex}}$
Graph each solution.
Graph numbers that make either inequality true.
	The solution covers all real numbers.
	$\left(\text{−}\infty ,\infty \right)$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{5}x-7\le -1$ or $\frac{1}{3}\left(x+6\right)\ge -2.$

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 the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{4}x-3\le 3$ or $\frac{2}{5}\left(x+10\right)\ge 0.$

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.

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?

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 $24.72+1.54x.$

Solve the inequality.
Answer the question.	The property owner can use 21–95 hcf and still fall within the “normal usage” billing range.

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 ?52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?

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

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 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?

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 $x<b.$
  
  $\begin{array}{cccc}\text{Other forms:}\hfill & & & \begin{array}{ccccccccccccc}a<x<b\hfill & & & \text{is equivalent to}\hfill & & & a<x\hfill & & & \text{and}\hfill & & & x<b\hfill \\ a\le x\le b\hfill & & & \text{is equivalent to}\hfill & & & a\le x\hfill & & & \text{and}\hfill & & & x\le b\hfill \\ a>x>b\hfill & & & \text{is equivalent to}\hfill & & & a>x\hfill & & & \text{and}\hfill & & & x>b\hfill \\ a\ge x\ge b\hfill & & & \text{is equivalent to}\hfill & & & a\ge x\hfill & & & \text{and}\hfill & & & x\ge b\hfill \end{array}\hfill \end{array}$
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.

Practice Makes Perfect

Solve Compound Inequalities with “and”

In the following exercises, solve each inequality, graph the solution, and write the solution in interval notation.

$x<3$ and $x\ge 1$

$x\le 4$ and $x>-2$

The solution is negative 2 is less than x which is less than or equal to 4. Its graph has an open circle at 1negative 2 and a closed circle at 4 with shading between the open and closed circles. Its interval notation is negative 2 to 4 within a parenthesis and a bracket.

$x\ge -4$ and $x\le -1$

$x>-6$ and $x<-3$

The solution is negative 6 is less than x which is less than negative 3. Its graph has an open circle at negative 6 and an open circle at negative 3 with shading between open circles. Its interval notation is negative 6 to negative 3 within parentheses.

$5x-2<8$ and $6x+9\ge 3$

$4x-1<7$ and $2x+8\ge 4$

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

$4x+6\le 2$ and

$2x+1\ge -5$

$4x-2\le 4$ and

$7x-1>-8$

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

$2x-11<5$ and

$3x-8>-5$

$7x-8<6$ and

$5x+7>-3$

The solution is negative 2 is less than x which is less than 2. Its graph has an open circle at negative 2 and an open circle at 2 with shading between the open circles. Its interval notation is negative 2 to 2 within parentheses.

$4\left(2x-1\right)\le 12$ and

$2\left(x+1\right)<4$

$5\left(3x-2\right)\le 5$ and

$3\left(x+3\right)<3$

The solution is x is less than negative 2. Its graph has an open circle at negative 2 and is shaded to the left. Its interval notation is negative infinity to negative 2 within parentheses.

$3\left(2x-3\right)>3$ and

$4\left(x+5\right)\ge 4$

$-3\left(x+4\right)<0$ and

$-1\left(3x-1\right)\le 7$

The solution is x is greater than or equal to negative 2. Its graph has a closed circle at negative 2 and is shaded to the right. Its interval notation is negative 2 to infinity within a bracket and a parenthesis.

$\frac{1}{2}\left(3x-4\right)\le 1$ and

$\frac{1}{3}\left(x+6\right)\le 4$

$\frac{3}{4}\left(x-8\right)\le 3$ and

$\frac{1}{5}\left(x-5\right)\le 3$

The solution is x is less than or equal to 12. Its graph has a closed circle at 12 and is shaded to the left. Its interval notation is negative infinity to 12 within a parenthesis and a bracket.

$5x-2\le 3x+4$ and

$3x-4\ge 2x+1$

$\frac{3}{4}x-5\ge -2$ and

$-3\left(x+1\right)\ge 6$

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

$\frac{2}{3}x-6\ge -4$ and

$-4\left(x+2\right)\ge 0$

$\frac{1}{2}\left(x-6\right)+2<-5$ and

$4-\frac{2}{3}x<6$

$-5\le 4x-1<7$

$-3<2x-5\le 1$

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 and is shaded between the open and closed circles. Its interval notation is 1 to 3 within a parenthesis and a bracket.

$5<4x+1<9$

$-1<3x+2<8$

The solution is negative 1 is less than x which is less than 2. Its graph has an open circle at negative 1 an open circle at 2 and is shaded between. Its interval notation is negative 1 to 2 within parentheses.

$-8<5x+2\le -3$

$-6\le 4x-2<-2$

The solution is negative 1 is less than or equal to x which is less than or 0. Its graph has a closed circle at negative 1 and an open circle at 0 and is shaded between the closed and open circles. Its interval notation is negative 1 to 0 within a bracket and a parenthesis.

Solve Compound Inequalities with “or”

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$x\le -2$ or $x>3$

$x\le -4$ or $x>-3$

The solution is x is less than or equal to negative 4 or x is greater than negative 3. The graph of the solutions on a number line has a closed circle at negative 4 and shading to the left and an open circle at negative 3 with shading to the right. The interval notation is the union of negative infinity to negative 4 within a parenthesis and a bracket and negative 3 and infinity within parentheses.

$x<2$ or $x\ge 5$

$x<0$ or $x\ge 4$

The solution is x is less than 0 or x is greater than or equal to 2. The graph of the solutions on a number line has an open circle at 0 and shading to the left and a closed circle at 4 with shading to the right. The interval notation is the union of negative infinity to 0 within parentheses and 4 to infinity within a bracket and parenthesis.

$2+3x\le 4$ or

$5-2x\le -1$

$4-3x\le -2$ or

$2x-1\le -5$

The solution is x is less than or equal to negative 2 or x is greater than or equal to 2. The graph of the solutions on a number line has a closed circle at negative 2 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 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

$2\left(3x-1\right)<4$ or

$3x-5>1$

$3\left(2x-3\right)<-5$ or

$4x-1>3$

The solution is x is less than two-thirds or x is greater than 1. The graph of the solutions on a number line has an open circle at two-thirds and shading to the left and an open circle at 1 with shading to the right. The interval notation is the union of negative infinity to two-thirds within parentheses and 1 and infinity within parentheses.

$\frac{3}{4}x-2>4$ or $4\left(2-x\right)>0$

$\frac{2}{3}x-3>5$ or $3\left(5-x\right)>6$

The solution is x is less than 3 or x is greater than 12. The graph of the solutions on a number line has an open circle at 3 and shading to the left and an open circle at 12 with shading to the right. The interval notation is the union of negative infinity to 3 within parentheses and 12 and infinity within parentheses.

$3x-2>4$ or $5x-3\le 7$

$2\left(x+3\right)\ge 0$ or

$3\left(x+4\right)\le 6$

$\frac{1}{2}x-3\le 4$ or

$\frac{1}{3}\left(x-6\right)\ge -2$

$\frac{3}{4}x+2\le -1$ or

$\frac{1}{2}\left(x+8\right)\ge -3$

Mixed practice

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$3x+7\le 1$ and

$2x+3\ge -5$

$6\left(2x-1\right)>6$ and

$5\left(x+2\right)\ge 0$

The solution is x is less than 1. Its graph has an open circle at negative 1 is shaded to the right. Its interval notation is 1 to infinity within parentheses.

$4-7x\ge -3$ or

$5\left(x-3\right)+8>3$

$\frac{1}{2}x-5\le 3$ or

$\frac{1}{4}\left(x-8\right)\ge -3$

$-5\le 2x-1<7$

$\frac{1}{5}\left(x-5\right)+6<4$ and

$3-\frac{2}{3}x<5$

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

$4x-2>6$ or

$3x-1\le -2$

$6x-3\le 1$ and

$5x-1>-6$

The solution is negative 1 is less than x which is less than or equal to two-thirds. Its graph has an open circle at negative 1 and a closed circle at two-thirds and is shaded between the open and closed circles. Its interval notation is negative 1 to two-thirds within a parenthesis and a bracket.

$-2\left(3x-4\right)\le 2$ and

$-4\left(x-1\right)<2$

$-5\le 3x-2\le 4$

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 a closed circle at 2 and is shaded between the closed circles. Its interval notation is negative 1 to 4 within brackets.

Solve Applications with Compound Inequalities

In the following exercises, solve.

Penelope is playing a number game with her sister June. Penelope is thinking of a number and wants June to guess it. Five more than three times her number is between 2 and 32. Write a compound inequality that shows the range of numbers that Penelope might be thinking of.

Gregory is thinking of a number and he wants his sister Lauren to guess the number. His first clue is that six less than twice his number is between four and forty-two. Write a compound inequality that shows the range of numbers that Gregory might be thinking of.

$5\le n\le 24$

Andrew is creating a rectangular dog run in his back yard. The length of the dog run is 18 feet. The perimeter of the dog run must be at least 42 feet and no more than 72 feet. Use a compound inequality to find the range of values for the width of the dog run.

Elouise is creating a rectangular garden in her back yard. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than 48 feet. Use a compound inequality to find the range of values for the width of the garden.

$6\le w\le 12$

Everyday Math

Blood Pressure A person’s blood pressure is measured with two numbers. The systolic blood pressure measures the pressure of the blood on the arteries as the heart beats. The diastolic blood pressure measures the pressure while the heart is resting.

ⓐ Let x be your systolic blood pressure. Research and then write the compound inequality that shows you what a normal systolic blood pressure should be for someone your age.

ⓑ Let y be your diastolic blood pressure. Research and then write the compound inequality that shows you what a normal diastolic blood pressure should be for someone your age.

Body Mass Index (BMI) is a measure of body fat is determined using your height and weight.

ⓐ Let x be your BMI. Research and then write the compound inequality to show the BMI range for you to be considered normal weight.

ⓑ Research a BMI calculator and determine your BMI. Is it a solution to the inequality in part (a)?

ⓐ answers vary ⓑ answers vary

Writing Exercises

In your own words, explain the difference between the properties of equality and the properties of inequality.

Explain the steps for solving the compound inequality $2-7x\ge -5$ or $4\left(x-3\right)+7>3.$

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve compound inequalities with “and.” In row 3, the I can was solve compound inequalities with “or.” In row 4, the I can was solve applications with compound inequalities.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Glossary

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

License

Icon for the Creative Commons Attribution 4.0 International License

Intermediate Algebra imported into JK's account for testing at Douglas College Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.