warm up solve. 3 x + 2 = 8 2. 3. -2k + 7 = -3
DESCRIPTION
Warm Up Solve. 3 x + 2 = 8 2. 3. -2k + 7 = -3. Objectives. Identify solutions of inequalities with one variable. Write and graph inequalities with one variable. . ≥. ≤. ≠. A < B. A > B. A ≤ B. A ≥ B. A ≠ B. A is greater than or equal to B . A is less - PowerPoint PPT PresentationTRANSCRIPT
Holt Algebra 1
3-1 Graphing and Writing InequalitiesWarm Up
Solve.1. 3x + 2 = 8
2.
3. -2k + 7 = -3
Holt Algebra 1
3-1 Graphing and Writing Inequalities
Identify solutions of inequalities with one variable.
Write and graph inequalities with one variable.
Objectives
Holt Algebra 1
3-1 Graphing and Writing Inequalities
An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs:
≤A ≤ BA is less than or
equal to B.
<A < BA is lessthan B.
>A > BA is greater
than B.
≥A ≥ B
A is greaterthan or
equal to B.
≠A ≠ BA is not
equal to B.
A solution of an inequality is any value that makes the inequality true.
Holt Algebra 1
3-1 Graphing and Writing InequalitiesExample 1: Identifying Solutions of Inequalities
Describe the solutions of x – 6 ≥ 4 in words.
When the value of x is a number less than 10, the value of x – 6 is less than 4. When the value of x is 10, the value of x – 6 is equal to 4.
When the value of x is a number greater than 10, the value of x – 6 is greater than 4.It appears that the solutions of x – 6 ≥ 4 are all real numbers greater than or equal to 10.
Solution?–9 4≥
?
–3–9
No–6 4≥
? 3.9 4 ≥ ? 4 4≥
? 4.1 4≥ ? 6 4≥
?
xx – 6
x – 6 ≥ 4
0 9.9 10 10.1 12–6 3.9 4 4.1 6
NoNo Yes Yes Yes
?
Holt Algebra 1
3-1 Graphing and Writing Inequalities
An inequality like 3 + x < 9 has too many solutions to list. You can use a graph on a number line to show all the solutions.
The solutions are shaded and an arrow shows that the solutions continue past those shown on the graph. To show that an endpoint is a solution, draw a solid circle at the number. To show an endpoint is not a solution, draw an empty circle.
Holt Algebra 1
3-1 Graphing and Writing Inequalities
Holt Algebra 1
3-1 Graphing and Writing InequalitiesExample 2: Graphing Inequalities
Graph each inequality.A. m ≥
0 1– 2 3 3
B. t < 5(–1 + 3) t < 5(–1 + 3) t < 5(2) t < 10
–4 –2 0 2 4 6 8 10 12 –6 –8
Holt Algebra 1
3-1 Graphing and Writing InequalitiesExample 3: Writing an Inequality from a Graph
Write the inequality shown by each graph.
Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2 means that 2 is not a solution, so use <.
x < 2
Use any variable. The arrow points to the right, so use either > or ≥. The solid circle at –0.5 means that –0.5 is a solution, so use ≥.
x ≥ –0.5
Holt Algebra 1
3-1 Graphing and Writing Inequalities
Reading Math “No more than” means “less than or equal to.” “At least” means “greater than or equal to”.
Holt Algebra 1
3-1 Graphing and Writing InequalitiesExample 4: Application
Ray’s dad told him not to turn on the air conditioner unless the temperature is at least 85°F. Define a variable and write an inequality for the temperatures at which Ray can turn on the air conditioner. Graph the solutions.Let t represent the temperatures at which Ray can turn on the air conditioner.
75 80 85 9070
Turn on the AC when temperature is at least 85°Ft ≥ 85
Draw a solid circle at 85. Shade all numbers greater than 85 and draw an arrow pointing to the right.
t 85
Holt Algebra 1
3-1 Graphing and Writing Inequalities
A store’s employees earn at least $8.50 per hour. Define a variable and write an inequality for the amount the employees may earn per hour. Graph the solutions.
Check It Out! Example 4
Let w represent an employee’s wages.
An employee earns at least $8.50w ≥ 8.50
4 6 8 10 12−2 0 2 14 16 188.5
w ≥ 8.5
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Warm-Up1. Describe the solutions of 7 < x + 4.
2. Graph h ≥ –4.75
–5 –4.75 –4.5Write the inequality shown by each graph.
3.
4.5. A cell phone plan offers free minutes for no more
than 250 minutes per month. Define a variable and write an inequality for the possible number of
free minutes. Graph the solution.
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Solve one-step inequalities by using addition.Solve one-step inequalities by using subtraction.
Objectives
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Example 1A: Using Addition and Subtraction to SolveInequalities
Solve the inequality and graph the solutions.
x + 12 < 20 x + 12 < 20
–12 –12x + 0 < 8
x < 8
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
d – 5 > –7
+5 +5d + 0 > –2
d > –2
d – 5 > –7
Example 1B: Using Addition and Subtraction to SolveInequalities
Solve the inequality and graph the solutions.
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Example 1C: Using Addition and Subtraction to SolveInequalities
Solve the inequality and graph the solutions.
0.9 ≥ n – 0.3
0 1 2
+0.3 +0.31.2 ≥ n – 0
1.2 ≥ n
0.9 ≥ n – 0.3
1.2
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Example 2: Problem-Solving ApplicationSami has a gift card. She has already used $14 of the of the total value, which was $30. Write, solve, and graph an inequality to show how much more she can spend.
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Make a Plan
Example 2 Continued
Write an inequality.Let g represent the remaining amount of money Sami can spend.
g + 14 ≤ 30
Amount remaining plus $30.is at
mostamount
used
g + 14 ≤ 30
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Solve
g + 14 ≤ 30– 14 – 14
g + 0 ≤ 16g ≤ 16
0 2 4 6 8 10 12 14 16 18 10
Example 2 Continued
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Check It Out! Example 2The Recommended Daily Allowance (RDA) of iron for a female in Sarah’s age group (14-18 years) is 15 mg per day. Sarah has consumed 11 mg of iron today. Write and solve an inequality to show how many more milligrams of iron Sarah can consume without exceeding RDA.
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Make a Plan
Write an inequality.Let x represent the amount of iron Sarah needs to consume.
Amount taken plus 15 mgis at
mostamount needed
11 + x 15
11 + x 15
Check It Out! Example 2 Continued
Holt Algebra 1
3-2 Solving Inequalities by Adding or Subtracting
Solve
11 + x 15
x 4
0 1 2 3 4 5 6 7 8 9 10
x 4. Sarah can consume 4 mg or less of iron without exceeding the RDA.
Check It Out! Example 2 Continued
–11 –11
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Warm-Up
Solve each inequality and graph the solutions.
1. 13 < x + 7x > 6
2. –6 + h ≥ 15h ≥ 21
3. 6.7 + y ≤ –2.1y ≤ –8.8
4. A certain restaurant has room for 120 customers. On one night, there are 72 customers dining. Write and solve an inequality to show how many more people can eat at the restaurant.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Solve one-step inequalities by using multiplication.Solve one-step inequalities by using division.
Objectives
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Remember, solving inequalities is similar to solving equations. To solve an inequality that contains multiplication or division, undo the operation by dividing or multiplying both sides of the inequality by the same number.
The following rules show the properties of inequality for multiplying or dividing by a positive number. The rules for multiplying or dividing by a negative number appear later in this lesson.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Example 1A: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions. 7x > –42
7x > –42>
1x > –6 x > –6
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
3(2.4) ≤ 3
7.2 ≤ m (or m ≥ 7.2)
0 2 4 6 8 10 12 14 16 18 20
Example 1B: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
r < 16
0 2 4 6 8 10 12 14 16 18 20
Example 1C: Multiplying or Dividing by a Positive Number
Solve the inequality and graph the solutions.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
If you multiply or divide both sides of an inequality by a negative number, the resulting inequality is not a true statement. You need to reverse the inequality symbol to make the statement true.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Caution! Do not change the direction of the inequality symbol just because you see a negative sign. For example, you do not change the symbol when solving 4x < –24.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Example 2A: Multiplying or Dividing by a Negative Number
Solve the inequality and graph the solutions.–12x > 84
x < –7
–10 –8 –6 –4 –2 0 2 4 6–12–14–7
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
16 18 20 22 2410 14 26 28 3012
Example 2B: Multiplying or Dividing by a Negative Number
Solve the inequality and graph the solutions.
24 x (or x 24)
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Example 3: Application
$4.30 times number of tubes is at most $20.00.
4.30 • p ≤ 20.00
Jill has a $20 gift card to an art supply store where 4 oz tubes of paint are $4.30 each after tax. What are the possible numbers of tubes that Jill can buy?Let p represent the number of tubes of paint that Jill can buy.
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
4.30p ≤ 20.00
p ≤ 4.65…
Since p is multiplied by 4.30, divide both sides by 4.30. The symbol does not change.
Since Jill can buy only whole numbers of tubes, she can buy 0, 1, 2, 3, or 4 tubes of paint.
Example 3 Continued
Holt Algebra 1
3-3 Solving Inequalities by Multiplying or Dividing
Warm-UpSolve each inequality and graph the solutions.1. 8x < –24 x < –3 2. –5x ≥ 30 x ≤ –6
3. x > 20 4. x ≥ 6
5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy?0, 1, 2, 3, 4, 5, 6, or 7 shirts
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Warm-UpSolve each inequality and graph the solutions.
1. 8x < –24 x < –3 2. –5x ≥ 30 x ≤ –6
3. x > 20 4. x ≥ 6
5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy?0, 1, 2, 3, 4, 5, 6, or 7 shirts
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Solve inequalities that contain more than one operation.
Objective
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Inequalities that contain more than one operation require more than one step to solve. Use inverse operations to undo the operations in the inequality one at a time.
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step InequalitiesExample 1A: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
45 + 2b > 6145 + 2b > 61
–45 –452b > 16
b > 80 2 4 6 8 10 12 14 16 18 20
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
8 – 3y ≥ 298 – 3y ≥ 29–8 –8
–3y ≥ 21
y ≤ –7–10 –8 –6 –4 –2 0 2 4 6 8 10
–7
Example 1B: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 1cSolve the inequality and graph the solutions.
1 – 2n ≥ 21–1 –1
–2n ≥ 20
n ≤ –10
–10–20 –12 –8 –4–16 0
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides by using the order of operations, combining like terms, or using the Distributive Property.
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Example 2A: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.2 – (–10) > –4t
12 > –4t
–3 < t (or t > –3)
–3–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Example 2B: Simplifying Before Solving InequalitiesSolve the inequality and graph the solutions.
–4(2 – x) ≤ 8
−4(2 – x) ≤ 8−4(2) − 4(−x) ≤
–8 + 4x ≤ 8+8 +8
4x ≤ 16
x ≤ 4
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 2bSolve the inequality and graph the solutions.
3 + 2(x + 4) > 33 + 2(x + 4) > 3
3 + 2x + 8 > 32x + 11 > 3
– 11 – 112x > –8
x > –4
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 2cSolve the inequality and graph the solutions.
5 < 3x – 2+2 + 2
7 < 3x
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 2c ContinuedSolve the inequality and graph the solutions.
7 < 3x
4 6 82 100
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Example 3: ApplicationTo rent a certain vehicle, Rent-A-Ride charges $55.00 per day with unlimited miles. The cost of renting a similar vehicle at We Got Wheels is $38.00 per day plus $0.20 per mile. For what number of miles is the cost at Rent-A-Ride less than the cost at We Got Wheels?
Let m represent the number of miles. The cost for Rent-A-Ride should be less than that of We Got Wheels.
Cost at Rent-A-Ride
must be less than
daily cost at We Got Wheels
plus$0.20
per mile times # of miles.
55 < 38 + 0.20 m
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
85 < m
Since 38 is added to 0.20m, subtract 8 from both sides to undo the addition.
Since m is multiplied by 0.20, divide both sides by 0.20 to undo the multiplication.
Rent-A-Ride costs less when the number of miles is more than 85.
Example 3 Continued
55 < 38 + 0.20m
–38 –3855 < 38 + 0.20m
17 < 0.20m
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 3
The average of Jim’s two test scores must be at least 90 to make an A in the class. Jim got a 95 on his first test. What grades can Jim get on his second test to make an A in the class?Let x represent the test score needed. The average score is the sum of each score divided by 2.
First test score
plus second test score
divided by
number of scores
is greater than or equal to
total score
(95 + x) 2 ≥ 90
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Check It Out! Example 3 Continued
The score on the second test must be 85 or higher.
Since 95 is added to x, subtract 95 from both sides to undo the addition.
95 + x ≥ 180–95 –95
x ≥ 85
Since 95 + x is divided by 2, multiply both sides by 2 to undo the division.
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Lesson Quiz: Part ISolve each inequality and graph the solutions.1. 13 – 2x ≥ 21 x ≤ –4
2. –11 + 2 < 3p p > –3
3. 23 < –2(3 – t) t > 7
4.
Holt Algebra 1
3-4 Solving Two-Step and Multi-Step Inequalities
Lesson Quiz: Part II
5. A video store has two movie rental plans. Plan A includes a $25 membership fee plus $1.25 for each movie rental. Plan B costs $40 for unlimited movie rentals. For what number of movie rentals is plan B less than plan A? more than 12 movies
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Warm UpSolve each equation. 1. 2x = 7x + 15 2.
5. Solve and graph 5(2 – b) > 52.
3. 2(3z + 1) = –2(z + 3) 4. 3(p – 1) = 3p + 2
x = –3
b < –3
–5 –3 –2 –1–4 0–6
3y – 21 = 4 – 2y y = 5z = –1
no solution
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Solve inequalities that contain variable terms on both sides.
Objective
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 1A: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.y ≤ 4y + 18y ≤ 4y + 18–y –y0 ≤ 3y + 18
–18 – 18–18 ≤ 3y
–6 ≤ y (or y –6)
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
4m – 3 < 2m + 6–2m – 2m2m – 3 < + 6 + 3 + 32m < 9
4 5 6
Example 1B: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 2: Business Application
The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding. How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean?
Let w be the number of windows.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 2 Continued
312 + 12 • w < 36 • w312 + 12w < 36w
– 12w –12w312 < 24w
13 < wThe Home Cleaning Company is less expensive for houses with more than 13 windows.
To collect the variable terms, subtract 12w from both sides.
Since w is multiplied by 24, divide both sides by 24 to undo the multiplication.
HomeCleaningCompany
siding charge
plus$12 per window
# of windows
is lessthan
PowerClean
cost per window
# ofwindows.times
times
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Check It Out! Example 2A-Plus Advertising charges a fee of $24 plus $0.10 per flyer to print and deliver flyers. Print and More charges $0.25 per flyer. For how many flyers is the cost at A-Plus Advertising less than the cost of Print and More?
Let f represent the number of flyers printed.
24 + 0.10 • f < 0.25 • f
plus$0.10per
flyer
is lessthan
# of flyers.
A-PlusAdvertisingfee of $24
Print and More’s cost
per flyer
# of flyerstimes times
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Check It Out! Example 2 Continued
24 + 0.10f < 0.25f–0.10f –0.10f
24 < 0.15f
160 < f
To collect the variable terms, subtract 0.10f from both sides.
Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication.
More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Check It Out! Example 3aSolve the inequality and graph the solutions.5(2 – r) ≥ 3(r – 2)
5(2 – r) ≥ 3(r – 2)5(2) – 5(r) ≥ 3(r) + 3(–2)
10 – 5r ≥ 3r – 6+6 +616 − 5r ≥ 3r
+ 5r +5r16 ≥ 8r
–6 –2 0 2–4 4
16 ≥ 8r
2 ≥ r
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
There are special cases of inequalities called identities and contradictions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both SidesExample 4A: Identities and Contradictions
Solve the inequality.2x – 7 ≤ 5 + 2x2x – 7 ≤ 5 + 2x
–2x –2x–7 ≤ 5
The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
2(3y – 2) – 4 ≥ 3(2y + 7)
2(3y – 2) – 4 ≥ 3(2y + 7)2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
6y – 4 – 4 ≥ 6y + 216y – 8 ≥ 6y + 21
Example 4B: Identities and ContradictionsSolve the inequality.
–6y –6y
–8 ≥ 21No values of y make the inequality true. There are no solutions.
Holt Algebra 1
3-6 Solving Compound InequalitiesWarm-Up
Solve each inequality and graph the solutions.1. t < 5t + 24
2. 5x – 9 ≤ 4.1x – 813. 4b + 4(1 – b) > b – 9
Solve each inequality.
4. 2y – 2 ≥ 2(y + 7)
5. 2(–6r – 5) < –3(4r + 2)
Holt Algebra 1
3-6 Solving Compound Inequalities
Solve compound inequalities with one variable. Graph solution sets of compound inequalities with one variable.
Objectives
Holt Algebra 1
3-6 Solving Compound Inequalities
The inequalities you have seen so far are simple inequalities. When two simple inequalities are combined into one statement by the words AND or OR, the result is called a compound inequality.
Holt Algebra 1
3-6 Solving Compound Inequalities
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 1: Chemistry Application
The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions.Let p be the pH level of the shampoo.6.0 is less than
or equal topH level is less than
or equal to 6.5
6.0 ≤ p ≤ 6.56.0 ≤ p ≤ 6.5
5.9 6.1 6.2 6.36.0 6.4 6.5
Holt Algebra 1
3-6 Solving Compound Inequalities
In this diagram, oval A represents some integer solutions of x < 10 and oval B represents some integer solutions of x > 0. The overlapping region represents numbers that belong in both ovals. Those numbers are solutions of both x < 10 and x > 0.
Holt Algebra 1
3-6 Solving Compound Inequalities
You can graph the solutions of a compound inequality involving AND by using the idea of an overlapping region. The overlapping region is called the intersection and shows the numbers that are solutions of both inequalities.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 2A: Solving Compound Inequalities Involving
ANDSolve the compound inequality and graph the solutions.
–5 < x + 1 < 2 –5 < x + 1 < 2 –1 – 1 – 1–6 < x < 1
–10 –8 –6 –4 –2 0 2 4 6 8 10
Graph –6 < x.
Graph x < 1.Graph the intersection by
finding where the two graphs overlap.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 2B: Solving Compound Inequalities Involving
ANDSolve the compound inequality and graph the solutions.
8 < 3x – 1 ≤ 11
8 < 3x – 1 ≤ 11+1 +1 +1
9 < 3x ≤ 12
3 < x ≤ 4
Holt Algebra 1
3-6 Solving Compound Inequalities
–5 –4 –3 –2 –1 0 1 2 3 4 5
Graph 3 < x.
Graph x ≤ 4.
Graph the intersection by finding where the two graphs overlap.
Example 2B Continued
Holt Algebra 1
3-6 Solving Compound Inequalities
Solve the compound inequality and graph the solutions.
Check It Out! Example 2a
–9 < x – 10 < –5
+10 +10 +10–9 < x – 10 < –5
1 < x < 5
–5 –4 –3 –2 –1 0 1 2 3 4 5
Graph 1 < x.
Graph x < 5.
Graph the intersection by finding where the two graphs overlap.
Holt Algebra 1
3-6 Solving Compound Inequalities
In this diagram, circle A represents some integer solutions of x < 0, and circle B represents some integer solutions of x > 10. The combined shaded regions represent numbers that are solutions of either x < 0 or x >10.
Holt Algebra 1
3-6 Solving Compound Inequalities
You can graph the solutions of a compound inequality involving OR by using the idea of combining regions. The combine regions are called the union and show the numbers that are solutions of either inequality.
>
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 3A: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.8 + t ≥ 7 OR 8 + t < 28 + t ≥ 7 OR 8 + t < 2–8 –8 –8 −8
t ≥ –1 OR t < –6
–10 –8 –6 –4 –2 0 2 4 6 8 10
Graph t ≥ –1.
Graph t < –6.
Graph the union by combining the regions.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 3B: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.4x ≤ 20 OR 3x > 21
4x ≤ 20 OR 3x > 21
x ≤ 5 OR x > 7
0 2 4 6 8 10
Graph x ≤ 5.
Graph x > 7.
Graph the union by combining the regions.
–10 –8 –6 –4 –2
Holt Algebra 1
3-6 Solving Compound Inequalities
Solve the compound inequality and graph the solutions.
Check It Out! Example 3a
2 +r < 12 OR r + 5 > 192 +r < 12 OR r + 5 > 19–2 –2 –5 –5
r < 10 OR r > 14
–4 –2 0 2 4 6 8 10 12 14 16
Graph r < 10.
Graph r > 14.
Graph the union by combining the regions.
Solve each simple inequality.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 4A: Writing a Compound Inequality from a
GraphWrite the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the compound inequality involves OR.On the left, the graph shows an arrow pointing left, so use either < or ≤. The solid circle at –8 means –8 is a solution so use ≤. x ≤ –8On the right, the graph shows an arrow pointing right, so use either > or ≥. The empty circle at 0 means that 0 is not a solution, so use >. x > 0The compound inequality is x ≤ –8 OR x > 0.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample 4B: Writing a Compound Inequality from a Graph
The shaded portion of the graph is between the values –2 and 5, so the compound inequality involves AND.The shaded values are on the right of –2, so use > or ≥. The empty circle at –2 means –2 is not a solution, so use >.m > –2The shaded values are to the left of 5, so use < or ≤. The empty circle at 5 means that 5 is not a solution so use <.m < 5The compound inequality is m > –2 AND m < 5 (or -2 < m < 5).
Write the compound inequality shown by the graph.
Holt Algebra 1
3-6 Solving Compound Inequalities
Lesson Quiz: Part I
1. The target heart rate during exercise for a 15 year-old is between 154 and 174 beats per minute inclusive. Write a compound inequality to show the heart rates that are within the target range. Graph the solutions.
154 ≤ h ≤ 174
Holt Algebra 1
3-6 Solving Compound Inequalities
Lesson Quiz: Part II
Solve each compound inequality and graph the solutions.2. 2 ≤ 2w + 4 ≤ 12
–1 ≤ w ≤ 4
3. 3 + r > −2 OR 3 + r < −7 r > –5 OR r < –10
Holt Algebra 1
3-6 Solving Compound Inequalities
Lesson Quiz: Part III
Write the compound inequality shown by each graph.
4.
x < −7 OR x ≥ 05.
−2 ≤ a < 4
Holt Algebra 1
3-6 Solving Compound Inequalities
Warm-Up
1. The target heart rate during exercise for a 15 year-old is between 154 and 174 beats per minute inclusive. Write a compound inequality to show the heart rates that are within the target range. Graph the solutions.
Solve each compound inequality and graph the solutions.2. 2 ≤ 2w + 4 ≤ 12
3. 3 + r > −2 OR 3 + r < −7
Holt Algebra 1
3-6 Solving Compound Inequalities
Solve inequalities in one variable involving absolute value expressions.
Objectives
Holt Algebra 1
3-6 Solving Compound Inequalities
When an inequality contains an absolute-value expression, it can be rewritten as a compound inequality. The inequality |x| < 5 describes all real number whose distance from 0 is less than 5 units. The solutions can be written as -5 < x < 5.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample: Solving Absolute Value Inequalities involving <.
Solve the inequality and graph the solutions.
|x| + 3 < 12
– 3 – 3
|x| < 9
|x + 4| ≤ 2
x + 4 ≤ 2 x + 4 ≥ -2 -4 -4 -4 -4
x ≤ -2 and x ≥ -6
Holt Algebra 1
3-6 Solving Compound Inequalities
The inequality |x| > 5 describes all real number whose distance from 0 is greater than 5 units. The solutions are all numbers less than -5 or greater than 5. The solutions can be written as x < -5 OR x > 5.
Holt Algebra 1
3-6 Solving Compound InequalitiesExample: Solving Absolute Value Inequalities involving >.
Solve the inequality and graph the solutions.
|x| - 20 > -13
+ 20 +20
|x| > 7
|x - 8|+5 ≥ 11
x - 8 ≥ 6 x - 8 ≤ -6 +8 +8 +8 +8
x ≥ 14 or x ≤ 2
- 5 -5|x – 8| ≥ 6
x < -7 or x > 7