chapter 10
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10-1 Solving Two-Step Equations
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve.
1. x + 12 = 35
2. 8x = 120
3. = 7
4. –34 = y + 56
x = 23
x = 15
y = 63
Course 3
10-1 Solving Two-Step Equations
y = –90
y9
Problem of the Day
x is an odd integer. If you triple x and then subtract 7, you get a prime number. What is x? (Hint: Think about what the prime number must be in order for x to be an odd.)
x = 3
Course 3
10-1 Solving Two-Step Equations
Learn to solve two-step equations.
Course 3
10-1 Solving Two-Step Equations
Sometimes more than one inverse operation is needed to solve an equation. Before solving, ask yourself, “What is being done to the variable, and in what order?” Then work backward to undo the operations.
Course 3
10-1 Solving Two-Step Equations
The mechanic’s bill to repair Mr. Wong’s car was $650. The mechanic charges $45 an hour for labor, and the parts that were used cost $443. How many hours did the mechanic work on the car?
Additional Example 1: Problem Solving Application
Course 3
10-1 Solving Two-Step Equations
Additional Example 1 Continued
11 Understand the Problem
The answer is the number of hours the mechanic worked on the car.
List the important information:
Let h represent the hours the mechanic worked.
• The parts cost $443.• The labor cost $45 per hour.• The total bill was $650.
Total bill = Parts + Labor
650 = 443 + 45hCourse 3
10-1 Solving Two-Step Equations
Think: First the variable is multiplied by 45, and then 443 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 443 from both sides of the equation, and then divide both sides of the new equation by 45.
22 Make a Plan
Additional Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
650 = 443 + 45h
Solve33
–443 –443 Subtract to undo the addition.
207 = 45h
4.6 = h
The mechanic worked for 4.6 hours on Mr. Wong’s car.
Additional Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
Divide to undo multiplication.207 45h45 45=
If the mechanic worked 4.6 hours, the labor would be $45(4.6) = $207. The sum of the parts and the labor would be $443 + $207 = $650.
Look Back44
Additional Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
The mechanic’s bill to repair your car was $850. The mechanic charges $35 an hour for labor, and the parts that were used cost $275. How many hours did the mechanic work on your car?
Try This: Example 1
Course 3
10-1 Solving Two-Step Equations
Try This: Example 1 Continued
11 Understand the Problem
The answer is the number of hours the mechanic worked on your car.
List the important information:
Let h represent the hours the mechanic worked.
• The parts cost $275.
• The labor cost $35 per hour.
• The total bill was $850.
Total bill = Parts + Labor
850 = 275 + 35hCourse 3
10-1 Solving Two-Step Equations
Think: First the variable is multiplied by 35, and then 275 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 275 from both sides of the equation, and then divide both sides of the new equation by 35.
22 Make a Plan
Try This: Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
850 = 275 + 35h
Solve33
–275 –275 Subtract to undo the addition.
575 = 35h
16.4 h
The mechanic worked for about 16.4 hours on your car.
Try This: Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
Divide to undo multiplication.575 35h35 35=
If the mechanic worked 16.4 hours, the labor would be $35(16.4) = $574. The sum of the parts and the labor would be $275 + $574 = $849.
Look Back44
Try This: Example 1 Continued
Course 3
10-1 Solving Two-Step Equations
Additional Example 2A: Solving Two-Step Equations
A. + 7 = 22
Solve.
Think: First the variable is divided by 3, and then 7 is added. To isolate the variable, subtract 7, and then multiply by 3.
– 7 – 7 Subtract to undo addition.
n3
+ 7 = 22n3
Multiply to undo division.3 = 3 15n3n = 45
Course 3
10-1 Solving Two-Step Equations
n3
= 15
Additional Example 2A Continued
Substitute 45 into the original equation.
Check + 7 = 22n3
? + 7 = 2245
315 + 7 = 22
?
Course 3
10-1 Solving Two-Step Equations
Additional Example 2B: Solving Two-Step Equations
B. 2.7 = –1.3m + 6.6
Think: First the variable is multiplied by –1.3, and then 6.6 is added. To isolate the variable, subtract 6.6, and then divide by –1.3.
2.7 = –1.3m + 6.6–6.6 –6.6 Subtract to undo addition.–3.9 = –1.3m
Divide to undo multiplication.
3 = m
–1.3 –1.3–3.9 = –1.3m
Course 3
10-1 Solving Two-Step Equations
Additional Example 2C: Solving Two-Step Equations
C. = 9y – 4 3
Think: First 4 is subtracted from the variable, and then the result is divided by 3. To isolate the variable, multiply by 3, and then add 4.
= 9y – 43
y – 4 = 27 + 4 + 4 Add to undo subtraction.
y = 31
3 · 3 · Multiply to undo division.= 9y – 43
Course 3
10-1 Solving Two-Step Equations
Try This: Example 2A
A. + 5 = 29
Solve.
Think: First the variable is divided by 4, and then 5 is added. To isolate the variable, subtract 5, and then multiply by 4.
– 5 – 5 Subtract to undo addition.
n4
+ 5 = 29n4
Multiply to undo division.4 = 4 24n4
n = 96Course 3
10-1 Solving Two-Step Equations
Substitute 96 into the original equation.
Check + 5 = 29n4
? + 5 = 2996
424 + 5 = 29
?
Try This: Example 2A Continued
Course 3
10-1 Solving Two-Step Equations
Try This: Example 2B
B. 4.8 = –2.3m + 0.2
Think: First the variable is multiplied by –2.3, and then 0.2 is added. To isolate the variable, subtract 0.2, and then divide by –2.3.
4.8 = –2.3m + 0.2–0.2 –0.2 Subtract to undo addition.4.6 = –2.3m
Divide to undo multiplication.
–2 = m
–2.3 –2.3 4.6 = –2.3m
Course 3
10-1 Solving Two-Step Equations
Try This: Example 2C
C. = 8y – 2 4
Think: First 2 is subtracted from the variable, and then the result is divided by 4. To isolate the variable, multiply by 4, and then add 2.
= 8y – 24
y – 2 = 32 + 2 + 2 Add to undo subtraction.
y = 34
4 · 4 · Multiply to undo division.= 8y – 24
Course 3
10-1 Solving Two-Step Equations
Solve.
1. – 3 = 10
2. 7y + 25 = –24
3. –8.3 = –3.5x + 13.4
4. = 3
5. The cost for a new cell phone plan is $39 per month plus a one-time start-up fee of $78. If you are charged $1014, how many months will the contract last?
Lesson Quiz
y = –7
x = –117
Insert Lesson Title Here
x = 6.2
y = 28
24 months
x–9
y + 5 11
Course 3
10-1 Solving Two-Step Equations
10-2 Solving Multistep Equations
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve.
1. 3x = 102
2. = 15
3. z – 100 = –1
4. 1.1 + 5w = 98.6
x = 34
y = 225
z = 99
Course 3
10-2 Solving Multistep Equations
w = 19.5
y15
Problem of the Day
Ana has twice as much money as Ben, and Ben has three times as much as Clio. Together they have $160. How much does each person have?Ana, $96; Ben, $48; Clio, $16
Course 3
10-2 Solving Multistep Equations
Learn to solve multistep equations.
Course 3
10-2 Solving Multistep Equations
To solve a complicated equation, you may have to simplify the equation first by combining like terms.
Course 3
10-2 Solving Multistep Equations
Solve.
8x + 6 + 3x – 2 = 37
Additional Example 1: Solving Equations That Contain Like Terms
11x + 4 = 37 Combine like terms.
– 4 – 4 Subtract to undo addition.11x = 33
x = 3
Divide to undo multiplication.3311
11x11
=
Course 3
10-2 Solving Multistep Equations
Check
Additional Example 1 Continued
8x + 6 + 3x – 2 = 37
8(3) + 6 + 3(3) – 2 = 37?
24 + 6 + 9 – 2 = 37?
37 = 37?
Course 3
10-2 Solving Multistep Equations
Substitute 3 for x.
Solve.
9x + 5 + 4x – 2 = 42
Try This: Example 1
13x + 3 = 42 Combine like terms.
– 3 – 3 Subtract to undo addition.13x = 39
x = 3
Course 3
10-2 Solving Multistep Equations
Divide to undo multiplication.3913
13x13
=
Check
Try This: Example 1 Continued
9x + 5 + 4x – 2 = 42
9(3) + 5 + 4(3) – 2 = 42?
27 + 5 + 12 – 2 = 42 ?
42 = 42?
Course 3
10-2 Solving Multistep Equations
Substitute 3 for x.
If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) to clear the fractions before you isolate the variable.
Course 3
10-2 Solving Multistep Equations
Solve.
A. + = –
Additional Example 2: Solving Equations That Contain Fractions
34
74
5n4
Multiply both sides by 4 to clear fractions, and then solve.
( ) ( )74
–3 4
5n4
4 + = 4
( ) ( ) ( )5n4
74
–3 44 + 4 = 4
5n + 7 = –3
Course 3
10-2 Solving Multistep Equations
Distributive Property.
Additional Example 2 Continued
5n + 7 = –3 – 7 –7 Subtract to undo addition.
5n = –10
5n5
–10 5
= Divide to undo multiplication.
n = –2
Course 3
10-2 Solving Multistep Equations
Insert Lesson Title Here
The least common denominator (LCD) is the smallest number that each of the denominators will divide into.
Remember!
Course 3
10-2 Solving Multistep Equations
Solve.
B. + – =
Additional Example 2B: Solving Equations That Contain Fractions
23
The LCD is 18.
x 2
7x9
17 9
18( ) + 18( ) – 18( ) = 18( )7x9
x2
17 9
23
14x + 9x – 34 = 12
23x – 34 = 12 Combine like terms.
( )x2
23
7x9
17 918 + – = 18
Course 3
10-2 Solving Multistep Equations
Distributive Property.
Multiply both sides by the LCD.
Additional Example 2B Continued
23x = 46
= 23x23
4623 Divide to undo multiplication.
x = 2
+ 34 + 34 Add to undo subtraction.
23x – 34 = 12 Combine like terms.
Course 3
10-2 Solving Multistep Equations
Additional Example 2B Continued
69
69=
?
Check
x 2
7x9
17 9
+ – = 23
23 Substitute 2 for x.7(2)
9 + – =(2) 2
17 9
?
23
149 + – =2
2 17 9
?
23
149 + – =
17 9
?1
The LCD is 9.69
149 + – =9
9 17 9
?
Course 3
10-2 Solving Multistep Equations
Solve.
A. + = –
Try This: Example 2A
14
54
3n4
Multiply both sides by 4 to clear fractions, and then solve.
( ) ( )54
–1 4
3n4
4 + = 4
( ) ( ) ( )3n4
54
–1 44 + 4 = 4
3n + 5 = –1
Course 3
10-2 Solving Multistep Equations
Distributive Property.
Try This: Example 2A Continued
3n + 5 = –1 – 5 –5 Subtract to undo addition.
3n = –6
3n3
–6 3
= Divide to undo multiplication.
n = –2
Course 3
10-2 Solving Multistep Equations
Solve.
B. + – =
Try This: Example 2B
13
The LCD is 9.
x 3
5x9
13 9
9( ) + 9( )– 9( ) = 9( )5x9
x3
13 9
13
5x + 3x – 13 = 3
8x – 13 = 3 Combine like terms.
( )x3
13
5x9
13 9 9 + – = 9( )
Course 3
10-2 Solving Multistep Equations
Distributive Property.
Multiply both sides by the LCD.
8x = 16
= 8x8
16 8 Divide to undo multiplication.
x = 2
+ 13 + 13 Add to undo subtraction.
8x – 13 = 3 Combine like terms.
Try This: Example 2B Continued
Course 3
10-2 Solving Multistep Equations
39
39=
?
Check
x 3
5x9
13 9
+ – = 13
13 Substitute 2 for x.5(2)
9 + – =(2) 3
13 9
?
13
109 + – =2
3 13 9
?
The LCD is 9.39
109 + – =6
9 13 9
?
Try This: Example 2B Continued
Course 3
10-2 Solving Multistep Equations
When Mr. and Mrs. Harris left for the mall, Mrs. Harris had twice as much money as Mr. Harris had. While shopping, Mrs. Harris spent $54 and Mr. Harris spent $26. When they arrived home, they had a total of $46. How much did Mr. Harris have when he left home?
Additional Example 3: Money Application
Let h represent the amount of money that Mr. Harris had when he left home. So Mrs. Harris had 2h when she left home.
h + 2h – 26 – 54 = 46Mr. Harris $+ Mrs. Harris $ – Mr. Harris spent – Mrs. Harris spent = amount left
Course 3
10-2 Solving Multistep Equations
Additional Example 3 Continued
3h – 80 = 46 Combine like terms.
+ 80 +80 Add 80 to both sides.
3h = 126
3h3
1263= Divide both sides by 3.
h = 42
Mr. Harris had $42 when he left home.
Course 3
10-2 Solving Multistep Equations
When Mr. and Mrs. Wesner left for the store, Mrs. Wesner had three times as much money as Mr. Wesner had. While shopping, Mr. Wesner spent $50 and Mrs. Wesner spent $25. When they arrived home, they had a total of $25. How much did Mr. Wesner have when he left home?
Try This: Example 3
Let h represent the amount of money that Mr. Wesner had when he left home. So Mrs. Wesner had 3h when she left home.
h + 3h – 50 – 25 = 25Mr. Wesner $ + Mrs. Wesner $ – Mr. Wesner spent – Mrs. Wesner spent = amount left
Course 3
10-2 Solving Multistep Equations
Try This: Example 3 Continued
4h – 75 = 25 Combine like terms.
+ 75 +75 Add 75 to both sides.
4h = 100
4h4
1004= Divide both sides by 4.
h = 25
Mr. Wesner had $25 when he left home.
Course 3
10-2 Solving Multistep Equations
Solve.
1. 6x + 3x – x + 9 = 33
2. –9 = 5x + 21 + 3x
3. + =
5. Linda is paid double her normal hourly rate for each hour she works over 40 hours in a week. Last week she worked 52 hours and earned $544. What is her hourly rate?
Lesson Quiz
x = –3.75
x = 3
Insert Lesson Title Here
x = 2858
x8
33 8
6x 7
4. – =2x21
2521
$8.50
x = 1 916
Course 3
10-2 Solving Multistep Equations
10-3 Solving Equations with Variables on Both Sides
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve.
1. 2x + 9x – 3x + 8 = 16
2. –4 = 6x + 22 – 4x
3. + = 5
4. – = 3
x = 1
x = -13
x = 34
Course 3
10-3 Solving Equations with Variables on Both Sides
27
x7 7
1
9x16
2x4
18
x = 50
Problem of the Day
An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle?22.5 in.
Course 3
10-3 Solving Equations with Variables on Both Sides
Learn to solve equations with variables on both sides of the equal sign.
Course 3
10-3 Solving Equations with Variables on Both Sides
Some problems produce equations that have variables on both sides of the equal sign.
Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
A. 4x + 6 = x
Additional Example 1A: Solving Equations with Variables on Both Sides
4x + 6 = x– 4x – 4x
6 = –3x
Subtract 4x from both sides.
Divide both sides by –3.
–2 = x
6–3
–3x–3=
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
B. 9b – 6 = 5b + 18
Additional Example 1B: Solving Equations with Variables on Both Sides
9b – 6 = 5b + 18– 5b – 5b
4b – 6 = 18
4b 4
24 4 =
Subtract 5b from both sides.
Divide both sides by 4.
b = 6
+ 6 + 6
4b = 24Add 6 to both sides.
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
C. 9w + 3 = 5w + 7 + 4w
Additional Example 1C: Solving Equations with Variables on Both Sides
9w + 3 = 5w + 7 + 4w
3 ≠ 7
9w + 3 = 9w + 7 Combine like terms.
– 9w – 9w Subtract 9w from both sides.
No solution. There is no number that can be substituted for the variable w to make the equation true.
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
A. 5x + 8 = x
Try This: Example 1A
5x + 8 = x– 5x – 5x
8 = –4x
Subtract 4x from both sides.
Divide both sides by –4.
–2 = x
8–4
–4x–4=
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
B. 3b – 2 = 2b + 123b – 2 = 2b + 12
– 2b – 2b
b – 2 = 12
Subtract 2b from both sides.
+ 2 + 2
b = 14Add 2 to both sides.
Try This: Example 1B
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w
1 ≠ 8
3w + 1 = 3w + 8 Combine like terms.
– 3w – 3w Subtract 3w from both sides.
No solution. There is no number that can be substituted for the variable w to make the equation true.
Try This: Example 1C
Course 3
10-3 Solving Equations with Variables on Both Sides
To solve multistep equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
A. 10z – 15 – 4z = 8 – 2z - 15
Additional Example 2A: Solving Multistep Equations with Variables on Both Sides
10z – 15 – 4z = 8 – 2z – 15
+ 15 +15
6z – 15 = –2z – 7 Combine like terms.+ 2z + 2z Add 2z to both sides.
8z – 15 = – 7
8z = 8
z = 1
Add 15 to both sides.
Divide both sides by 8.8z 88 8=
Course 3
10-3 Solving Equations with Variables on Both Sides
B.
Additional Example 2B: Solving Multistep Equations with Variables on Both Sides
Multiply by the LCD.
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14 Combine like terms.
y5
34
3y5
710
+ – = y –
y5
34
3y5
710
+ – = y –
20( ) = 20( )y5
34
3y5
710
+ – y –
20( ) + 20( ) – 20( )= 20(y) – 20( )y5
3y5
34
710
Course 3
10-3 Solving Equations with Variables on Both Sides
Additional Example 2B Continued
Add 14 to both sides.
–15 = 4y – 14
–1 = 4y
+ 14 + 14
–1 4
4y4 = Divide both sides by 4.
-14 = y
16y – 15 = 20y – 14
– 16y – 16y Subtract 16y from both sides.
Course 3
10-3 Solving Equations with Variables on Both Sides
Solve.
A. 12z – 12 – 4z = 6 – 2z + 32
Try This: Example 2A
12z – 12 – 4z = 6 – 2z + 32
+ 12 +12
8z – 12 = –2z + 38 Combine like terms.+ 2z + 2z Add 2z to both sides.
10z – 12 = + 38
10z = 50
z = 5
Add 12 to both sides.
Divide both sides by 10.10z 5010 10=
Course 3
10-3 Solving Equations with Variables on Both Sides
B.
Multiply by the LCD.
6y + 20y + 18 = 24y – 18
26y + 18 = 24y – 18 Combine like terms.
y4
34
5y6
68
+ + = y –
y4
34
5y6
68
+ + = y –
24( ) = 24( )y4
34
5y6
68
+ + y –
24( ) + 24( )+ 24( )= 24(y) – 24( )y4
5y6
34
68
Try This: Example 2B
Course 3
10-3 Solving Equations with Variables on Both Sides
Subtract 18 from both sides.
2y + 18 = – 18
2y = –36
– 18 – 18
–36 2
2y2 = Divide both sides by 2.
y = –18
26y + 18 = 24y – 18
– 24y – 24y Subtract 24y from both sides.
Try This: Example 2B Continued
Course 3
10-3 Solving Equations with Variables on Both Sides
Additional Example 3: Consumer Application
Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?
Course 3
10-3 Solving Equations with Variables on Both Sides
Additional Example 3 Continued
First solve for the price of one doughnut.
1.25 + 2d = 0.50 + 5dLet d represent the price of one doughnut.
– 2d – 2d
1.25 = 0.50 + 3dSubtract 2d from both sides.
– 0.50 – 0.50Subtract 0.50 from both sides.
0.75 = 3d
0.753
3d3= Divide both sides by 3.
0.25 = d The price of one doughnut is $0.25.
Course 3
10-3 Solving Equations with Variables on Both Sides
Additional Example 3 Continued
Now find the amount of money Jamie spends each morning.
1.25 + 2d Choose one of the original expressions.
Jamie spends $1.75 each morning.
1.25 + 2(0.25) = 1.75
0.25n0.25
1.75 0.25 =
Let n represent the number of doughnuts.
Find the number of doughnuts Jamie buys on Tuesday.
0.25n = 1.75
n = 7; Jamie bought 7 doughnuts on Tuesday.
Divide both sides by 0.25.
Course 3
10-3 Solving Equations with Variables on Both Sides
Try This: Example 3
Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?
Course 3
10-3 Solving Equations with Variables on Both Sides
Try This: Example 3 Continued
First solve for distance around the track.
2x + 4 = 4x + 2Let x represent the distance around the track.
– 2x – 2x
4 = 2x + 2Subtract 2x from both sides.
– 2 – 2 Subtract 2 from both sides.
2 = 2x
22
2x2= Divide both sides by 2.
1 = x The track is 1 mile around.
Course 3
10-3 Solving Equations with Variables on Both Sides
Try This: Example 3 Continued
Now find the total distance Helene walks each day.
2x + 4 Choose one of the original expressions.
Helene walks 6 miles each day.2(1) + 4 = 6
Let n represent the number of 1-mile laps.
Find the number of laps Helene walks on Saturdays.
1n = 6
Helene walks 6 laps on Saturdays.
n = 6
Course 3
10-3 Solving Equations with Variables on Both Sides
Lesson Quiz
Solve.
1. 4x + 16 = 2x
2. 8x – 3 = 15 + 5x
3. 2(3x + 11) = 6x + 4
4. x = x – 9
5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each?
x = 6
x = –8
Insert Lesson Title Here
no solution
x = 3614
12
An orange has 45 calories. An apple has 75 calories.
Course 3
10-3 Solving Equations with Variables on Both Sides
10-4 Solving Multistep Inequalities
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve.
1. 6x + 36 = 2x
2. 4x – 13 = 15 + 5x
3. 5(x – 3) = 2x + 3
4. + x =
x = –9
x = –28
x = 6
Course 3
10-4 Solving Multistep Inequalities
78
316
1116
x = –
Problem of the Day
Find an integer x that makes the following two inequalities true:4 < x2 < 16 and x < 2.5
x = –3
Course 3
10-4 Solving Multistep Inequalities
Learn to solve two-step inequalities and graph the solutions of an inequality on a number line.
Course 3
10-4 Solving Multistep Inequalities
Solving a multistep inequality uses the same inverse operations as solving a multistep equation. Multiplying or dividing the inequality by a negative number reverses the inequality symbol.
Course 3
10-4 Solving Multistep Inequalities
Solve and graph.
Additional Example 1A: Solving Multistep Inequalities
A. 4x + 1 > 13
4x + 1 > 13 – 1 – 1 Subtract 1 from both sides.
4x > 124x4
> 124
Divide both sides by 4.
x > 3 1 2 3 4 5 6 7
Course 3
10-4 Solving Multistep Inequalities
Additional Example 1B: Solving Multistep Inequalities
B. –7 < 3x + 8
–7 < 3x + 8
– 8 – 8 Subtract 8 from both sides.
–15 < 3x
– 15 3
< 3x 3 Divide both sides by 3.
–5 < x -7 -6 -5 -4 -3 -2 -1
Course 3
10-4 Solving Multistep Inequalities
Additional Example 1C: Solving Multistep Inequalities
C. -9x + 7 25
–9x + 7 25
– 7 – 7 Subtract 7 from both sides.
–9x 18
–9x–9
18–9
Divide each side by –9; change to .
x –2-6 -5 -4 -3 -2 -1 0
Course 3
10-4 Solving Multistep Inequalities
Solve and graph.
Try This: Example 1A
A. 5x + 2 > 12
5x + 2 > 12 – 2 – 2 Subtract 2 from both sides.
5x > 105x5
> 105
Divide both sides by 5.
x > 2 1 2 3 4 5 6 7
Course 3
10-4 Solving Multistep Inequalities
B. –5 < 2x + 9
–5 < 2x + 9
– 9 – 9 Subtract 9 from both sides.
–14 < 2x
– 14 2
< 2x 2 Divide both sides by 2.
–7 < x -7 -6 -5 -4 -3 -2 -1
Try This: Example 1B
Course 3
10-4 Solving Multistep Inequalities
C. -4x + 2 18
–4x + 2 18
– 2 – 2 Subtract 2 from both sides.
–4x 16
–4x–4
16–4
Divide each side by –4; change to .
x –4-6 -5 -4 -3 -2 -1 0
Try This: Example 1C
Course 3
10-4 Solving Multistep Inequalities
Solve and graph.
Additional Example 2A: Solving Multistep Inequalities
A. 10x + 21 – 4x < –15
10x + 21 – 4x < –15
– 21 – 21 Subtract 21 from both sides.
6x6
< –36 6 Divide both sides by 6.
x < –6-8 -7 -6 -5 -4 -3 -2
6x + 21 < –15 Combine like terms.
6x < –36
Course 3
10-4 Solving Multistep Inequalities
Additional Example 2B: Solving Multistep Inequalities
Multiply by LCD, 20.
8x + 15 18
– 15 – 15 Subtract 15 from both sides.
8x 3
B. + 2x5
34
910
+ 2x5
34
910
20( + ) 20( )2x5
34
910
20( ) + 20( ) 20( )2x5
34
910
Course 3
10-4 Solving Multistep Inequalities
Additional Example 2 Continued
x 38
8x8
38 Divide both sides by 8.
8x 3
Course 3
10-4 Solving Multistep Inequalities
0 1
38
Additional Example 2C: Solving Multistep Inequalities
C. 8x + 8 > 11x – 1
8x + 8 > 11x – 1– 8x – 8x Subtract 8x from both sides.
8 > 3x – 1
93
> 3x 3
Add 1 to each side.
3 > x
-1 0 1 2 3 4 5
+1 +1
9 > 3x
Course 3
10-4 Solving Multistep Inequalities
Divide both sides by 3.
Solve and graph.
Try This: Example 2A
A. 15x + 30 – 5x < –10
15x + 30 – 5x < –10
– 30 – 30 Subtract 30 from both sides.
10x10
< –40 10 Divide both sides by 10.
x < –4-8 -7 -6 -5 -4 -3 -2
10x + 30 < –10 Combine like terms.
10x < –40
Course 3
10-4 Solving Multistep Inequalities
Try This: Example 2B
Multiply by LCD, 20.
12x + 5 10
– 5 – 5 Subtract 5 from both sides.
12x 5
B. + 3x5
14
510
+ 3x5
14
510
20( + ) 20( )3x5
14
510
20( ) + 20( ) 20 ( )3x5
14
510
Course 3
10-4 Solving Multistep Inequalities
Try This: Example 2B Continued
x 512
12x12
512 Divide both sides by 12.
12x 5
0 5 12
Course 3
10-4 Solving Multistep Inequalities
Try This: Example 2C
C. 4x + 3 > 8x – 1
4x + 3 > 8x – 1– 4x – 4x Subtract 4x from both sides.
3 > 4x – 1
44
> 4x 4
Add 1 to each side.
1 > x
-1 0 1 2 3 4 5
+1 +1
4 > 4x
Course 3
10-4 Solving Multistep Inequalities
Divide both sides by 4.
Additional Example 3: Business Application
A school’s Spanish club is selling bumper stickers. They bought 100 bumper stickers for $55, and they have to give the company 15 cents for every sticker sold. If they plan to sell each bumper sticker for $1.25, how many do they have to sell to make a profit?Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost.
R > C
Course 3
10-4 Solving Multistep Inequalities
Additional Example 3 Continued
The revenue from selling x bumper stickers at $1.25 each is 1.25x. The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 55 + 0.15x. Substitute the expressions for R and C.
1.25x > 55 + 0.15x Let x represent the number of bumper stickers sold. Fixed cost is $55. Unit cost is 15 cents.
Course 3
10-4 Solving Multistep Inequalities
Additional Example 3 Continued
– 0.15x – 0.15x Subtract 0.15x from both sides.
1.10x > 55
x > 50
The Spanish club must sell more than 50 bumper stickers to make a profit.
Divide both sides by 1.10.
1.25x > 55 + 0.15x
1.10x1.10
551.10>
Course 3
10-4 Solving Multistep Inequalities
Try This: Example 3
R > C
A school’s Spanish club is selling bumper stickers. They bought 200 bumper stickers for $45, and they have to give the company 25 cents for every sticker sold. If they plan to sell each bumper sticker for $2.50, how many do they have to sell to make a profit?Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost.
Course 3
10-4 Solving Multistep Inequalities
Try This: Example 3 Continued
The revenue from selling x bumper stickers at $2.50 each is 2.5x. The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 45 + 0.25x. Substitute the expressions for R and C.
2.5x > 45 + 0.25x Let x represent the number of bumper stickers sold. Fixed cost is $45. Unit cost is 25 cents.
Course 3
10-4 Solving Multistep Inequalities
– 0.25x – 0.25x Subtract 0.25x from both sides.
2.25x > 45
x > 20
The Spanish club must sell more than 20 bumper stickers to make a profit.
Divide both sides by 2.25.
2.5x > 45 + 0.25x
2.25x2.25
452.25>
Try This: Example 3 Continued
Course 3
10-4 Solving Multistep Inequalities
Lesson Quiz: Part 1
Solve and graph.
1. 4x – 6 > 10
2. 7x + 9 < 3x – 15
3. w – 3w < 32
4. w +
x < –6
x > 4
Insert Lesson Title Here
w > –1623
14
12
w 38
1 2 3 4 5 6 7
-10 -9 -8 -7 -6 -5 -4
-18 -17 -16 -15 -14 -13 -12
0 38
Course 3
10-4 Solving Multistep Inequalities
Lesson Quiz: Part 2
5. Antonio has budgeted an average of $45 a month for entertainment. For the first five months of the year he has spent $48, $39, $60, $48, and $33. How much can Antonio spend in the sixth month without exceeding his average budget?
no more than $42
Course 3
10-4 Solving Multistep Inequalities
10-5 Solving for a Variable
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve.
1. 8x – 9 = 23
2. 9x + 12 = 4x + 37
3. 6x – 8 = 7x + 3
4. x + 3 =
x = 4
x = 5
x = –11
Course 3
10-5 Solving for a Variable
12
18
23 4x = – , or –5 3
4
Problem of the Day
The formula A = 4r2 gives the surface area of a geometric figure. Solve the formula for r.
Can you identify what the geometric figure is?sphere
Course 3
10-5 Solving for a Variable
Learn to solve an equation for a variable.
Course 3
10-5 Solving for a Variable
If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation.
Course 3
10-5 Solving for a Variable
Solve for the indicated variable.
Additional Example 1A: Solving for a Variable by Addition or Subtraction
A. Solve a – b + 1 = c for a.
a – b + 1 = c
+ b – 1 + b – 1 Add b and subtract 1 from both sides.
a = c + b – 1 Isolate a.
Course 3
10-5 Solving for a Variable
Solve for the indicated variable.
Additional Example 1B: Solving for a Variable by Addition or Subtraction
Course 3
10-5 Solving for a Variable
B. Solve a – b + 1 = c for b.
a – b + 1 = c– a – 1 – a – 1 Subtract a and 1 from both
sides.
–b = c – a – 1 Isolate b.
Multiply both sides by –1.–1 (–b) = –1 (c – a – 1)
b = –c + a + 1 Isolate b.
Solve for the indicated variable.
Try This: Example 1A
A. Solve y – b + 3 = c for y.
y – b + 3 = c
+ b – 3 + b – 3 Add b and subtract 3 from both sides.
y = c + b – 3 Isolate y.
Course 3
10-5 Solving for a Variable
Solve for the indicated variable.
Try This: Example 1B
Course 3
10-5 Solving for a Variable
B. Solve p – w + 4 = f for w.
p – w + 4 = f– p – 4 – p – 4 Subtract p and 4 from both
sides.
–w = f – p – 4 Isolate w.
Multiply both sides by –1.–1 (–w) = –1 (f – p – 4)
w = –f + p + 4 Isolate w.
Course 3
10-5 Solving for a Variable
To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values.
Solve for the indicated variable. Assume all values are positive.
Additional Example 2A: Solving for a Variable by Division or Square Roots
A. Solve A = s2 for s.
A = s2
Course 3
10-5 Solving for a Variable
√A = s Isolate s.
√A = √s2 Take the square root of both sides.
Additional Example 2B: Solving for a Variable by Division or Square Roots
B. Solve V = IR for R.
V = IR
= Divide both sides by I.IRI
VI
IV= R Isolate R.
Course 3
10-5 Solving for a Variable
Solve for the indicated variable. Assume all values are positive.
Additional Example 2C: Solving for a Variable by Division or Square Roots
C. Solve the formula for the area of a trapezoid for h. Assume all values are positive.
A = h(b1 + b2)12 Write the formula.
2 • A = 2 • h(b1 + b2)12 Multiply both sides by 2.
2A = h(b1 + b2)
Course 3
10-5 Solving for a Variable
2A(b1 + b2)
= h Isolate h.
2A(b1 + b2)
h(b1 + b2) (b1 + b2)
= Divide both sides by (b1 + b2).
Solve for the indicated variable. Assume all values are positive.
Try This: Example 2A
A. Solve w = g2 + 4 for g.
w = g2 + 4
Course 3
10-5 Solving for a Variable
√ w – 4 = g Isolate g.
√w – 4 = √g2
w – 4 = g2
– 4 – 4 Subtract 4 from both sides.
Take the square root of both sides.
B. Solve A = lw for w.
A = lw
= Divide both sides by l.lwl
Al
lA= w Isolate w.
Course 3
10-5 Solving for a Variable
Solve for the indicated variable. Assume all values are positive.
Try This: Example 2B
C. Solve s = 180(n – 2) for n.
s = 180(n – 2)
Course 3
10-5 Solving for a Variable
Solve for the indicated variable. Assume all values are positive.
Try This: Example 2C
s 180(n – 2)180 180
= Divide both sides by 180.
s180
= (n – 2)
Add 2 to both sides.
s180
+ 2 = n
+ 2 + 2
Course 3
10-5 Solving for a Variable
To find solutions (x, y), choose values for x substitute to find y.
Remember!
Additional Example 3: Solving for y and Graphing
Solve for y and graph 3x + 2y = 8.
3x + 2y = 8
–3x –3x
2y = –3x + 8
–3x + 82
2y2
=
y = + 4–3x2
x y
–2 7
0 4
2 1
4 –2
Course 3
10-5 Solving for a Variable
Additional Example 3 Continued
Course 3
10-5 Solving for a Variable
3x + 2y = 8
Try This: Example 3
Solve for y and graph 4x + 3y = 12.
4x + 3y = 12
–4x – 4x
3y = –4x + 12
–4x + 12 3
3y3
=
y = + 4–4x3
x y
–3 8
0 4
3 0
6 –4
Course 3
10-5 Solving for a Variable
Try This: Example 3 Continued
x
y
Course 3
10-5 Solving for a Variable
–4 –2 2 4 6
10
86
2
–2
–4
44x + 3y = 12
–6
Lesson Quiz: Part 1
Solve for the indicated variable.
1. P = R – C for C.
2. P = 2l+ 2w for l.
3. V = Ah for h.
4. R = for S.
C = R - P
Insert Lesson Title Here
C – Rt = S
13C – S
t
= h3VA
= lP – 2w2
Course 3
10-5 Solving for a Variable
Lesson Quiz: Part 2
5. Solve for y and graph 2x + 7y = 14.
Insert Lesson Title Here
y = – + 2 2x 7
Course 3
10-5 Solving for a Variable
10-6 Systems of Equations
Course 3
Warm UpWarm Up
Problem of the DayProblem of the Day
Lesson PresentationLesson Presentation
Warm UpSolve for the indicated variable.
1. P = R – C for R
2. V = Ah for A
3. R = for C
R = P + C
Rt + S = C
Course 3
10-6 Systems of Equations
13C – S
t
= h3VA
Problem of the Day
At an audio store, stereos have 2 speakers and home-theater systems have 5 speakers. There are 30 sound systems with a total of 99 speakers. How many systems are stereo systems and how many are home-theater systems?
17 stereo systems, 13 home-theater systems
Course 3
10-6 Systems of Equations
Learn to solve systems of equations.
Course 3
10-6 Systems of Equations
Vocabulary
system of equationssolution of a system of equations
Insert Lesson Title Here
Course 3
10-6 Systems of Equations
A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs.
Course 3
10-6 Systems of Equations
Determine if the ordered pair is a solution of the system of equations below.
5x + y = 7x – 3y = 11
Additional Example 1A: Identifying Solutions of a System of Equations
A. (1, 2)
5x + y = 7
5(1) + 2 = 7?
7 = 7
x – 3y = 11
1 – 3(2) = 11 ? Substitute for
x and y.–5 11
The ordered pair (1, 2) is not a solution of the system of equations.
Course 3
10-6 Systems of Equations
Additional Example 1B: Identifying Solutions of a System of Equations
B. (2, –3)
5(2) + –3 = 7 ?
7 = 7
2 – 3(–3) = 11? Substitute for
x and y.11 = 11
The ordered pair (2, –3) is a solution of the system of equations.
Determine if the ordered pair is a solution of the system of equations below.
5x + y = 7x – 3y = 11
5x + y = 7 x – 3y = 11
Course 3
10-6 Systems of Equations
Additional Example 1C: Identifying Solutions of a System of Equations
C. (20, 3)
5(20) + (3) = 7 ?
103 7
20 – 3(3) = 11? Substitute for
x and y.11 = 11
The ordered pair (20, 3) is not a solution of the system of equations.
Determine if the ordered pair is a solution of the system of equations below.
5x + y = 7x – 3y = 11
5x + y = 7 x – 3y = 11
Course 3
10-6 Systems of Equations
Determine if each ordered pair is a solution of the system of equations below.
4x + y = 8x – 4y = 12
Try This: Example 1A
A. (1, 2)
4x + y = 8
4(1) + 2 = 8?
6 8
x – 4y = 12
1 – 4(2) = 12 ? Substitute for
x and y.–7 12
The ordered pair (1, 2) is not a solution of the system of equations.
Course 3
10-6 Systems of Equations
Try This: Example 1B
Determine if each ordered pair is a solution of the system of equations below.
4x + y = 8x – 4y = 12
B. (2, –3)
4(2) + –3 = 8 ?
5 8
2 – 4(–3) = 12? Substitute for
x and y.14 12
The ordered pair (2, –3) is not a solution of the system of equations.
4x + y = 8 x – 4y = 12
Course 3
10-6 Systems of Equations
Try This: Example 1C
C. (1, 4)
The ordered pair (1, 4) is not a solution of the system of equations.
Determine if each ordered pair is a solution of the system of equations below.
4x + y = 8x – 4y = 12
4(1) + 4 = 8 ?
8 = 8
1 – 4(4) = 12? Substitute for
x and y.–15 12
4x + y = 8 x – 4y = 12
Course 3
10-6 Systems of Equations
Course 3
10-6 Systems of Equations
When solving systems of equations, remember to find values for all of the variables.
Helpful Hint
Additional Example 2: Solving Systems of Equations
Solve the system of equations. y = x – 4y = 2x – 9
Solve the equation to find x.
x – 4 = 2x – 9– x – x Subtract x from both sides.
–4 = x – 9
5 = x
+ 9 + 9 Add 9 to both sides.
y = x – 4 y = 2x – 9
y = y
x – 4 = 2x – 9
Course 3
10-6 Systems of Equations
Additional Example 2 Continued
To find y, substitute 5 for x in one of the original equations.
y = x – 4 = 5 – 4 = 1
The solution is (5, 1).
Check: Substitute 5 for x and 1 for y in each equation.
y = x – 4 y = 2x – 9
1 = 5 – 4? 1 = 2(5) – 9
?
1 = 1 1 = 1
Course 3
10-6 Systems of Equations
Try This: Example 2
Solve the system of equations. y = x – 5y = 2x – 8
Solve the equation to find x.
x – 5 = 2x – 8– x – x Subtract x from both sides.
–5 = x – 8
3 = x
+ 8 + 8 Add 8 to both sides.
y = x – 5 y = 2x – 8
y = y
x – 5 = 2x – 8
Course 3
10-6 Systems of Equations
Try This: Example 2 Continued
To find y, substitute 3 for x in one of the original equations.
y = x – 5 = 3 – 5 = –2
The solution is (3, –2).
Check: Substitute 3 for x and –2 for y in each equation.
y = x – 5 y = 2x – 8
–2 = 3 – 5 ? –2 = 2(3) – 8
?
–2 = –2 –2 = –2
Course 3
10-6 Systems of Equations
To solve a general system of two equations with two variables, you can solve both equations for x or both for y.
Course 3
10-6 Systems of Equations
Additional Example 3A: Solving Systems of Equations
Solve the system of equations.
A. x + 2y = 8 x – 3y = 13x + 2y = 8 x – 3y = 13
–2y –2y + 3y + 3y
Solve both equations for x.
x = 8 – 2y x = 13 + 3y
8 – 2y = 13 + 3y+ 2y + 2y
8 = 13 + 5y
Add 2y to both sides.
Course 3
10-6 Systems of Equations
Additional Example 3A Continued
8 = 13 + 5y
–13 –13
–5 = 5y
Subtract 13 from both sides.
–55
5y 5 = Divide both sides
by 5.–1 = y
x = 8 – 2y = 8 – 2(–1) Substitute –1 for y. = 8 + 2 = 10The solution is (10, –1).
Course 3
10-6 Systems of Equations
Course 3
10-6 Systems of Equations
You can choose either variable to solve for. It is usually easiest to solve for a variable that has a coefficient of 1.
Helpful Hint
Additional Example 3B: Solving Systems of Equations
Solve the system of equations.
B. 3x – 3y = -3 2x + y = -53x – 3y = –3 2x + y = –5
–3x –3x –2x –2x
Solve both equations for y.
–3y = –3 – 3x y = –5 – 2x
–3–3
3x–3
–3y–3 = –
y = 1 + x
1 + x = –5 – 2x
Course 3
10-6 Systems of Equations
Additional Example 3B Continued
+ 2x + 2x Add 2x to both sides.
1 + 3x = –5–1 –1
3x = –6
1 + x = –5 – 2x
Subtract 1 from both sides.
–6 3
3x3 =
Divide both sides by 3.
x = –2y = 1 + x
= 1 + –2 = –1 Substitute –2 for x.
The solution is (–2, –1).
Course 3
10-6 Systems of Equations
Try This: Example 3A
Solve the system of equations.
A. x + y = 5 3x + y = –1x + y = 5 3x + y = –1
–x –x – 3x – 3x
Solve both equations for y.
y = 5 – x y = –1 – 3x
5 – x = –1 – 3x+ x + x
5 = –1 – 2x
Add x to both sides.
Course 3
10-6 Systems of Equations
Try This: Example 3A Continued
5 = –1 – 2x
+ 1 + 1
6 = –2x
Add 1 to both sides.
Divide both sides by –2.
–3 = x
y = 5 – x = 5 – (–3) Substitute –3 for x. = 5 + 3 = 8The solution is (–3, 8).
Course 3
10-6 Systems of Equations
Try This: Example 3B
Solve the system of equations.
B. x + y = –2 –3x + y = 2x + y = –2 –3x + y = 2
– x – x + 3x + 3x
Solve both equations for y.
y = –2 – x y = 2 + 3x
–2 – x = 2 + 3x
Course 3
10-6 Systems of Equations
+ x + x Add x to both sides.
–2 = 2 + 4x–2 –2
–4 = 4x
–2 – x = 2 + 3x
Subtract 2 from both sides.
Divide both sides by 4.–1 = x
y = 2 + 3x= 2 + 3(–1) = –1 Substitute –1
for x.The solution is (–1, –1).
Course 3
10-6 Systems of Equations
Try This: Example 3B Continued
Lesson Quiz
1. Determine if the ordered pair (2, 4) is a solution of the system. y = 2x; y = –4x + 12
Solve each system of equations.
2. y = 2x + 1; y = 4x
3. 6x – y = –15; 2x + 3y = 5
4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers.
yes
Insert Lesson Title Here
(–2,3)
15 and 8
( , 2)12
Course 3
10-6 Systems of Equations