Solving and Graphing One Variable Compound Inequalities

LHS-1

2x < 4

.LHS /2.<.RHS /2.

x < 2

The solution set of this inequality can be graphed by marking every number to the left of 2, not including 2.

Similarly, to solve the second inequality, 5 needs to be added to both sides of the inequality.

5x - 5 > 0

▼

Solve for x

AddIneq

LHS+5>RHS+5

5x > 5

.LHS /5.>.RHS /5.

x > 1

Therefore, the solution set of this inequality are the numbers to the right of 1 in a number line, not including 1.

What would happen if these two inequalities were combined? For combined inequalities, x might solve both inequalities at the same time or either inequality at any given time. How would these solution sets look? These kinds of combined inequalities arise in all kinds of situations. Because of this, it is important to study them.

Discussion

Compound Inequalities

Combining two or more inequalities with the word and or or yields what is called a compound inequality.

Compound Inequality	Is Read As
x < 5 or x > 8	x is less than 5 or greater than 8.
x > 2 and x ≤ 4	x is greater than 2 and less than or equal to 4.

Compound inequalities using the word and are commonly written without showing the actual word. Consider the following example. x > 2 and x ≤ 4 The first inequality can be rewritten as 2 < x because the statement x is greater than 2 is equivalent to 2 is less than x. With this change, the inequality can be rewritten as follows. 2 < x and x ≤ 4 ⇒ 2 < x ≤ 4

It is important to note that compound inequalities written with the variable between such symbols, similar to the given example are compound inequalities pertaining to and.

Method

Solving a Compound Inequality

Solving a compound inequality is done by first separating each individual inequality. Then, each inequality is solved one at a time in the regular fashion. Lastly, the solution sets are combined. As an example, consider the following compound inequality. - 3 < 2x - 1 ≤ 2 This inequality will be solved to illustrate each step of this method.

Separate the Compound Inequality

Before a compound inequality can be solved, each individual inequality has to be identified. - 3 < 2x - 1 ≤ 2 ⇓ - 3 < 2x - 1 and 2x - 1 ≤ 2

Solve the Individual Inequalities

The individual inequalities can now be solved one at a time. For the example used, - 3 < 2x - 1 will be solved first, using inverse operations.

- 3 < 2x - 1

AddIneq

LHS+1

- 3 + 1 < 2x - 1 + 1

AddTerms

Add terms

- 2 < 2x

.LHS /2.<.RHS /2.

- 2/2 < 2x/2

SimpQuot

Simplify quotient

- 1 < x

The solution set is - 1 < x. Next is 2x - 1 ≤ 2, which is solved in a similar manner.

2x - 1 ≤ 2

AddIneq

LHS+1≤RHS+1

2x - 1 + 1 ≤ 2 + 1

AddTerms

Add terms

2x ≤ 3

.LHS /2.≤.RHS /2.

2x/2 ≤ 3/2

SimpQuot

Simplify quotient

x ≤ 3/2

The solution set of the second inequality is all values of x that satisfy x ≤ 32.

Combine the Solution Sets

Lastly, combining the two solution sets yields the solution set of the compound inequality. Here, the compound inequality was written in the condensed form of an and inequality. Therefore, the two solution sets can written separately with the word and, or they can be combined as follows. - 1 < x and x ≤ 32 ⇓ - 1 < x ≤ 3/2

Concept

Solution Set of a Compound Inequality

The solution set of a compound inequality consists of the solution sets of the individual inequalities. For compound inequalities with or, a solution of either individual inequality is a solution of the compound inequality. Consider the following compound inequality. x ≥ 2 or x < - 2 The graph of this compound inequality is the union of the graphs of the individual inequalities. These graphs are recognized by the fact that they continue infinitely in either direction.

A solution of a compound inequality with and, however, must be a solution of both individual inequalities. Consider the following inequality. x < 1 and x ≥ - 3 The graph of the compound inequality is the intersection of the graphs of the individual inequalities.

Unlike the graphs of compound inequalities with or, compound inequalities written with and do not always extend infinitely.

Example

The Speed of a Tractor

Kevin's class is having a work experience internship this week. He is working with a farmer to test the speed of an autonomous tractor in a field.

In the first test, the tractor started from a point 2 miles away from the barn. After a half hour, the tractor was at least 10 miles away from the barn. In the next test, the tractor started at a point 1.5 miles away from the barn. When it stopped 45 minutes later, it was less than 20 miles from the barn.

a Considering the test results, what are the possible speeds in miles per hour at which the tractor could drive? Round the answer to one decimal place.

b Graph the possible speeds on a number line.

Answer

a 16 ≤ r < 24.7

Hint

a Write the given information as inequalities.

b Graph the individual inequalities, then combine the graphs.

Solution

a To find the possible speeds, both test needs to be considered. It is given that in the first test, the tractor started 2 miles away from the barn and stopped at least 10 miles away from the barn after half an hour. This means that the sum of the distance traveled and the starting distance of 2 miles has to be greater than or equal to 10 miles.

distance traveled + 2 ≥ 10 To find the distance traveled, a modification of the speed formula will be used. r = d/t In this formula, r is the speed or rate, d is the distance traveled, and t is the time it took to travel the distance. To find the distance, d needs to be the isolated variable. r = d/t ⇒ d = rt Therefore, this formula can be substituted for the distance traveled in the written inequality. rt + 2 ≥ 10 Since the speed is asked in miles per hour and the tractor drove for half an hour, the given time is written as 12=0.5. 0.5r + 2 ≥ 10 Likewise in the second test, it is given that the tractor starts from 1.5 miles away and ends less than 20 miles away from the barn after 45 minutes. This means that the sum of the distance traveled and 1.5 miles has to be less than 20 miles. distance traveled + 1.5 < 20 Then, the formula for the distance traveled can be substituted in this inequality as well. rt + 1.5 < 20 In this case 45 minutes can be converted to hours using a conversion factor. 45 minutes are equal to 4560 = 0.75 hours. 0.75r + 1.5 < 20 Since the possible values for the speed are the values of r that satisfy both inequalities, these inequalities can be combined as a compound inequality with and. 0.5r + 2 ≥ 10 and 0.75r + 1.5 < 20 To solve this compound inequality, the properties of inequalities will be used. To solve the inequality on the left, first 2 will be subtracted from both sides of the inequality.

0.5r + 2 ≥ 10

▼

Solve for r

LHS-2≥RHS-2

0.5r ≥ 8

.LHS /0.5.≥.RHS /0.5.

r ≥ 16

Similarly, to solve the inequality on the right, 1.5 will be subtracted from both sides of the inequality.

0.75r + 1.5 < 20

▼

Solve for r

LHS-1.5

0.75r < 18.5

.LHS /0.75.<.RHS /0.75.

r < 24.666666...

RoundDec

Round to 1 decimal place(s)

r < 24.7

The compound inequality can be written with the found solutions. r ≥ 16 and r < 24.7 It should be noted that some compound inequalities that are written with and can be rewritten as follows. 16 ≤ r < 24.7 This means that the possible speeds of the tractor are greater than or equal to 16 miles per hour and less than 24.7 miles per hour.

b To graph the solution set of the compound inequality, it is useful to graph the solution set of each individual inequality first. Since the inequality is non-strict, the solution set of the inequality r ≥ 16 is made of all the numbers to the right of 16, including 16. This is shown by using a closed circle at 16.

On this case, since the inequality is strict, the solution set of the inequality r < 24.7 is made of the numbers to the left of 24.7. Since 24.7 is not included in the solution set, an open circle is used instead.

Since the compound inequality is written with and, its solution set is made of the numbers that satisfy both inequalities.

Example

Securing the Area Around a Volcano

Ignacio is interning at the local disaster preparedness center. He is using a simulator to help to secure the unsafe area near a volcano that is about to erupt. He is in charge of marking safe distances to the east and the west of the volcano. He initially marked the safe distance with flags, one 30 miles to the east of the base of the volcano and the other 15 miles to the west.

As time passes and data comes in, Ignacio realizes that his estimates were wrong. He notes that the flag to the west is less than half the distance away from the volcano that it should be. On the other hand, he calculates that the eastern flag covers less than two thirds of the actual necessary safe distance from the volcano.

a Consider that the distance to the east of the volcano are represented by positive numbers and that distances to the west are represented by negative numbers. Write a compound inequality that describes the safe distances from the volcano.

b Graph these lengths on a number line.

Answer

a d < -30 or d > 45

Hint

a Rewrite the given information as individual inequalities.

b Graph the individual inequalities, then combine the graphs.

Solution

a To write the safe distance as a compound inequality, the individual inequalities must first be identified. It is given that the western flag is less than half the actual necessary safe distance. This can be written as an inequality, with d representing the total safe distance.

15 miles west<1/2d Since the distance is to the west of the volcano, it is written as a negative number. Also, since this value is negative, further distances from the volcano are less than -15. Therefore, the inequality has to be rewritten as follows. 1/2d < - 15 Similarly, it is given that the eastern flag is less than two thirds of the safe distance at 30 miles from the volcano. This can also be written as an inequality. Since the distances to the east are written with positive numbers, the inequality does not need to be modified. 30<2/3d These inequalities can be solved by using the Properties of Inequalities, a similar fashion to solving an equation. The first inequality is solved by multiplying by 2.

1/2d < - 15

▼

Simplify

MultIneq

LHS* 2 < RHS * 2

2/2d < 2(- 15)

SimpQuot

Simplify quotient

d < 2(- 15)

d < -30

The safe distance to the west is less than - 30 miles, or more than 30 miles. The second inequality can be solved by multiplying both sides by the reciprocal of 23, 32.

30 < 2/3d

▼

Simplify

MultIneq

LHS* 3/2 < RHS * 3/2

3/2* 30 < 3/2* 2/3d

MultFracByInverse

a/b* b/a=1

3/2* 30 < d

MoveRightFacToNum

a/c* b = a* b/c

3* 30/2 < d

90/2 < d

CalcQuot

Calculate quotient

45 < d

RearrangeIneq

Rearrange inequality

d > 45

The safe distance to the east is more than 45 miles. Since the safe distance can be either more than 30 miles west or more than 45 miles east, the compound inequality is written with or. d < -30 or d > 45

b To graph the solution set of a compound inequality, it is convenient to first graph each individual inequality. First, the solution set of d < -30 is made of all the numbers to the left of -30, not including -30 because the inequality is strict.

Similarly, the graph of the solution set of d > 45 is made of every number to the right of 45, not including 45.

Finally, since it is written with or, the graph of the compound inequality is the combination of both solution sets and does not need to be adjusted or limited.

Example

Eat or Save Cookies?

Tearrik is spending his week of work experience at a local bakery. On his first day, he bought 100 cookies at a discount. He decides to eat 5 cookies a day until they are all gone. Tearrik is also allowed to bring home 3 cookies per day, which he gives to his brother. Tearrik's brother decides that he will not eat his cookies until he has at least 90 saved up.

a Let d be the number of days that have passed since Tearrik bought his cookies. Write a compound inequality to find on which days both brothers will eat cookies together.

b Graph the solution set of the inequality set in Part A.

Answer

a d < 20 and d ≥ 30

Hint

a Rewrite the given information as individual inequalities.

b Graph the individual inequalities, then combine the graphs.

Solution

a To find on which days both brothers will eat cookies together, it is important to identify when each brother will eat cookies. Tearrik starts with 100 cookies and he eats 5 cookies every day until he reaches 0 cookies. This situation can be written as an inequality.

100 - 5d >0 Since Tearrik will eat cookies throughout this time, the values of d are the possible numbers of days in which he eats cookies. On the other hand, Tearrik's brother will save 3 cookies every day until he has at least 90 cookies. Then, Tearrik's brother will start eating his cookies. 3d ≥ 90 For both to eat cookies, the value d must satisfy both inequalities at the same time. This kind of compound inequality is written with and. 100-5d>0 and 3d ≥ 90 To solve this compound inequality, both inequalities must be solved. First, to solve the inequality on the left, 100 will be subtracted from both sides of the inequality.

100 - 5d >0

▼

Solve for d

LHS-100>RHS-100

- 5d >- 100

DivNegIneq

Divide by - 5 and flip inequality sign

d < 20

This means that Terrik will eat his cookies over the first 20 days. Since this inequality is strict, Tearrik will not reach the 20th day before he finishes his cookies. Then, the second inequality is solved by dividing both sides of the inequality by 3.

3d ≥ 90

.LHS /3.≥.RHS /3.

d ≥ 30

This means that Terrik's brother will save his cookies for 30 days. Since the inequality is non-strict, he will start eating his cookies on the 30th day. Now the compound inequality can be rewritten using the solutions. d < 20 and d ≥ 30 This means that Tearrik will be able to eat cookies for 20 days, while his brother will save his cookies for 30 days before eating any of them. Therefore, they will not eat any cookies on the same day.

b To graph the solution set of the compound inequality, first it is necessary to graph the solution set of each individual inequality. The solution set of the inequality d < 20 are the numbers less than 20, not including 20. Since 20 is not included in the solution set, it is marked with an open circle.

Similarly, the solution set of the inequality d ≥ 30 is made of the numbers greater than or equal to 30. Since thi inequality is not strict, the circle is closed.

The solution set of the compound inequality is made of the numbers that satisfy both inequalities at the same time. Since there are no such numbers, the compound inequality has no solution.

This confirms that the brothers will not eat cookies together.

Example

Different Thermometers for Different Temperatures

For his internship, Davontay is working in a research lab. He is researching what tools are used to measure really low and really high temperatures. He found out that a thermocouple thermometer can measure temperatures lower than 3272^(∘) F, while a pyrometer thermometer can measure temperatures greater than or equal to 973 K.

To help find these temperatures in degrees Celsius, the relationships between the different temperature scales are shown in the following table. It should be noted that C refers to a temperature in degrees Celsius, F is the temperature in degrees Fahrenheit, and K refers to kelvins.

Fahrenheit	Kelvin
9/5C + 32 = F	C + 273 = K

a What range of temperatures, in degrees Celsius, can be covered by using either thermometer?

b Graph the temperature range from Part A in a number line.

Answer

a C < 1800 or C ≥ 700

Hint

a Rewrite the given information as inequalities.

b Graph the individual inequalities, then combine the graphs.

Solution

a To find the range of temperatures in degrees Celsius that can be covered by the thermometers, first it is needed to write the given information as inequalities. It is given that a thermocouple can measure temperatures that are lower than 3272^(∘) F. This can be written as an inequality, with F as the temperature in degrees Fahrenheit.

F < 3272 Since the temperature is needed in degrees Celsius, the conversion formula will be substituted for F in the inequality above. Let C be the temperature in degrees Celsius. 9/5C + 32 < 3272 On the other hand, it is given that a pyrometer can measure temperatures as low as 973K. This can be written as an inequality, using K as the temperature in kelvins. It should be noted that kelvins are not degrees since it is an absolute scale. K ≥ 973 Substituting the conversion formula from the given table, the inequality can be rewritten in terms of degrees Celsius C. C + 273 ≥ 973 Since the question asks to describe the temperature range that both thermometers can measure, if either one can be used, the compound inequality is written with or. 9/5C + 32 < 3272 or C + 273 ≥ 973 Now, to solve this compound inequality, each inequality will be solved individually using the Properties of Inequalities. To solve the inequality on the left, first subtract 32 from both sides of the inequality.

9/5C + 32 < 3272

▼

Solve for C

LHS-32

9/5C < 3240

MultIneq

LHS* 5/9 < RHS * 5/9

C < 5/9 * 3240

MoveRightFacToNum

a/c* b = a* b/c

C < 5* 3240/9

C < 16200/9

CalcQuot

Calculate quotient

C < 1800

A similar process can be done to the inequality on the right.

C + 273 ≥ 973

LHS-273≥RHS-273

C ≥ 700

Therefore, the compound inequality can be rewritten with these solutions. C < 1800 or C ≥ 700

b To graph the temperature range from Part A, first the solution set of each individual inequality should be graphed. The solution set of the inequality C < 1800 is made of all the numbers to the left of 1800, not including 1800.

Similarly, the solution set of the inequality C ≥ 700 is made of every point to the right of 700, including 700.

Since the inequality is written with or, the solution set of the compound inequality is made of the numbers that satisfy either inequality. By combining the graphs it can be noted that every number is a solution to the compound inequality.

This means that any temperature can be measured using either of the thermometers.

Example

Saving for a Car

Diego is helping at a local car dealership for his work experience. After spending time at the dealership, he decides to start saving money to buy a car. His father told him that he would double the amount of money that Diego saves, starting from now. Also, Diego will receive extra 500 dollars from his uncle to help buy the car when he finishes saving.

When looking for prices, Diego notices that most of the cars he likes range from 15 thousand dollars to 18 thousand dollars.

a How could the situation be written as a compound inequality?

b How much does Diego need to save to afford the car?

c Graph the inequality set in Part B.

Answer

a 15 000 ≤ 2m + 500 ≤ 18 000

b Diego must save between $7250 and $8750, including these values, to afford the car.

Hint

a First write the total amount of money that Diego saves as an expression.

b Solve the compound inequality set in Part A.

c Graph each individual inequality, then combine the graphs.

Solution

a To write the situation as a compound inequality, first the total amount of money Diego needs to save should be expressed. Let m be the total amount of money that Diego saves. It is given that Diego's dad is going to double the amount of m. This means that m is multiplied by 2.

2m Also, Diego's uncle will give Diego $500 dollars after finishing saving. This adds 500 to the amount after it is multiplied by 2. With this information, it is possible to write Diego's total amount of money as an expression. 2m + 500 Diego needs from 15 thousand dollars to 18 thousand dollars to afford a car he likes. Therefore, the total amount of money that Diego needs must lie between these values in order for him to be able to afford the car. This can be expressed as a compound inequality. 2m + 500 ≥ 15 000 and 2m + 500 ≤ 18 000 Since the inequalities are written with and, they can be rewritten as follows. 15 000 ≤ 2m + 500 ≤ 18 000

b To find the amount of money that Diego himself needs to save to be able to afford the car, the compound inequality set in Part A needs to be solved for m. This can be done using the Properties of Inequalities to solving each individual inequality first.

15 000 ≤ 2m + 500

LHS-500≤RHS-500

14 500 ≤ 2m

.LHS /2.≤.RHS /2.

7250 ≤ m

Since the expression is the same, the second inequality is solved following the same steps.

2m + 500 ≤ 18 000

LHS-500≤RHS-500

2m ≤ 17 500

.LHS /2.≤.RHS /2.

m ≤ 8750

The resulting inequality is obtained by combining these results, similar to how it was written in Part A. 7250 ≤ m ≤ 8750 Since the inequality is non-strict, Diego can start looking to buy the car after he saves some amount of money from $7250 to $8750, including these values. After he saves enough, he can stop saving money.

c To graph the solution set of the compound inequality set in Part B, the solution set of each individual inequality will be graphed first. The solution set of m≥7250 is made of all the numbers to the right of 7250, including 7250.

Similarly, the graph of the solution set of inequality m≤ 8750 is made of all the points to the left of 8750, including 8750.

Finally, the solution set of the compound inequality is made of every number that both graphs share. In this case, the overlapping space from 7250 to 8750, inclusive.

Pop Quiz

Practice Solving Compound Inequalities

Various different compound inequalities will be shown in the applet below. Select the correct solution set.

Closure

Distances From Home to the Beach

At the beginning of the lesson, it was asked how far Vincenzo's house was from the beach. The following information was given.

Vincenzo will drive at 50 miles per hour.
He will arrive home in 40 to 55 minutes.

Then, the following questions were asked.

a How far can Vincenzo's house be from the beach? Round to one decimal place.

b Supposing Vincenzo's assumption is correct, let y be an invalid distance from his house to the beach. How could the possible values of y be written?

Answer

a Between 33.3 and 45.8 miles away.

b y < 33.3 or y > 45.8

Hint

a Write the situation as a compound inequality.

b Find the values that do not satisfy the compound inequality set in Part A.

Solution

a To find the distance from the beach to Vincenzo's house, a modification of the speed formula will be used.

r = d/t In this formula, r is the rate or speed, d is the distance traveled, and t is the time duration of the movement. Since two time estimations are given, the time has to be the isolated quantity in the formula.

r = d/t

▼

Solve for t

MultEqn

LHS * t=RHS* t

tr = d

DivEqn

.LHS /r.=.RHS /r.

t = d/r

It is given that it will take Vincenzo between 40 and 55 minutes to get home. Therefore, the time t in the formula has to be at least 40. This is the same as writing that t is greater than or equal to 40 minutes. t ≥ 40 minutes However, since the speed is given in miles per hour, the 40 minute interval must be written in terms of hours. This can be done using a conversion factor. 40 minutes* 1 hour/60 minutes = 2/3 hours The inequality can be rewritten using this information. t ≥ 2/3 hours Now the formula for the time in terms of the distance and the speed can be substituted into the inequality. It is given that Vincenzo will drive at 50 miles per hour. This means that r is equal to 50. d/50 ≥ 2/3 hours The same can be done with the upper limit of 55 minutes. The first to note is that the statement can be written as t is less than or equal to 55 minutes t ≤ 55 minutes Using the same conversion factor, the time can be rewritten in terms of hours. 55 minutes* 1 hour/60 minutes = 11/12 hours And using this time span, the inequality can be rewritten. t ≤ 11/12 hours Finally, the expression in terms of the distance can be substituted into the inequality. d/50 ≤ 11/12 hours Since the possible values for d satisfy both inequalities, the compound inequality is written with and. d/50 ≥ 2/3 and d/50 ≤ 11/12 Recall that compound inequalities written with and can be rewritten as follows. 2/3 ≤ d/50 ≤ 11/12 Finally, to find the possible values of d, each section of the compound inequality is multiplied by 50.

2/3 ≤ d/50 ≤ 11/12

▼

Solve for d

MultEqn

LHS * 50=RHS* 50

50* 2/3 ≤ d ≤ 50* 11/12