Solving Linear Equations and Inequalities

CHAPTER

2.1 Equations, Formulas, and the Problem-Solving Process

2.2 The Addition Principle of Equality2.3 The Multiplication Principle of Equality2.4 Applying the Principles to Formulas2.5 Translating Word Sentences to Equations2.6 Solving Linear Inequalities

22

Equations, Formulas, and the Problem-Solving Process2.12.1

1. Verify solutions to equations.2. Use formulas to solve problems.

Equation: Two expressions set equal.

For example, 4x + 5 = 9 is an equation made from the expressions 4x + 5 and 9.

Solution: A number that makes an equation true when it replaces the variable in the equation.

For example, the number 5 is the solution to the equation x + 2 = 7 because when 5 replaces x, it makes the equation true.

Slide 2- 3

Checking a Possible SolutionTo determine whether a value is a solution to a given equation, replace the variable in the equation with the value. If the resulting equation is true, the value is a solution.

Slide 2- 4

Example 1a

Check to see if –3 is a solution to 6x + 5 = –13.Solution

6x + 5 = –13 6(–3) + 5 = –13 –18 + 5 = –13

–13 = –13

Since –3 makes the equation true, it is a solution to 6x + 5 = –13.

Replace x with –3 to see if the equation is true.

Slide 2- 5

Example 1b

Check to see if –0.2 is a solution to x2 + 0.7 = 1.4.Solution

x2 + 0.7 = 1.4 (−0.2)2 + 0.7 = 1.4 0.04 + 0.7 = 1.4

0.74 ≠ 1.4

Because −0.2 does not make the equation true, it is not a solution to x2 + 0.7 = 1.4.

Replace x with –0.2 to see if the equation is true.

Slide 2- 6

Problem-Solving Outline1. Understand the problem. a. Read the question(s) (not the whole problem, just

the question at the end) and write a note to yourself about what it is you are to find.

b. Read the whole problem, underlining the key words.

c. If possible or useful, draw a picture, make a list or table or organize what is known and unknown, simulate the situation, or search for a related example problem.

2. Plan your solution by searching for a formula or using key words to translate to an equation.

Slide 2- 7

Problem-Solving Outline continued3. Execute the plan by solving the equation/formula.4. Answer the question. Look at the note about what

you were to find and make sure you answer that question. Include appropriate units.

5. Check results. a. Try finding the solution in a different way, reversing

the process, or estimating the answer and make sure the estimate and actual answer are reasonably close.

b. Make sure the answer is reasonable.

Slide 2- 8

Formula: An equation that describes a mathematical relationship.

Perimeter: The distance around a figure.Area: The total number of square units that fill a figure.Volume: The total number of cubic units that fill a space.Circumference: The distance around a circle.Radius: The distance from the center of a circle to any point on the circle.Diameter: The distance across a circle through its center.

Slide 2- 9

Using a FormulaTo use a formula:1. Replace the variables with the corresponding given

values.2. Solve for the missing value.

Slide 2- 10

Geometric Formulas—Plane FiguresSquare Rectangle Parallelogram Trapezoid

P = 4sA = s2

P = 2l + 2wA = lw

A = bh

Triangle Circle

1( )

2A h a b

1

2A bh

2

or 2C d C r

A r

s

s

w

l

h

b

h

b

a

r

dh

b

Slide 2- 11

Geometric Formulas--SolidsBox Pyramid Cylinder Cone

Sphere

34

3V r

2 2 2

V lwh

SA lw lh wh

1

3V lwh

2V r h 21

3V r h

Slide 2- 12

L

W

Example 2a

A kennel owner is enclosing a rectangular area with fencing. The kennel is to be 60 feet in length and 25 feet wide. What is the total amount of fencing needed, disregarding a gate?

Understand The fencing creates the perimeter of the kennel.

Plan Because the shape is a rectangle, we can use the formula P = 2l + 2w.

Execute Replace l with 60 feet and w with 25 feet, then calculate.

Slide 2- 13

continued

Answer The fencing needed for the kennel is 170 ft.Check The solution can be verified using an

alternative method. Add all four side lengths.

2 2 P l w 2 60 2 25 P

120 50 P170 ft.P

60 60 25 25 170 ft. P

Slide 2- 14

Example 2b

A circular window is to be covered with an applied stencil. If the window has a radius of 3 feet, how much of the window cover is needed?

Understand The amount of the window stencil is the same as the area of the window.

Plan Because the shape is a circle, we can use the formula

Execute Replace with 3.14 and r with 3 ft. and simplify.

2.A r

Slide 2- 15

continued

Answer About 28.26 square feet of window stencil is needed to cover the window.

Check Verify the reasonableness of the answer using estimation. If we round to 3, the answer is (3)(3 ft)(3 ft) = 27 ft2. Because is a little more than 3 and our answer is a little more than 27 ft.2, our answer is reasonable.

2A r 23.14 3ft.A

23.14 9ft.A 228.26 ftA

Slide 2- 16

Calculating the Area of Composite FiguresFor composite figures:1. To calculate the area of a figure composed of two

or more figures that are next to each other, add the areas of the individual figures.

2. To calculate the area of a region defined by a smaller figure within a larger figure, subtract the area of the smaller figure from the area of the larger figure.

Slide 2- 17

Example 3

Following is a drawing of a deck that is to have brick flooring installed. Calculate the area of the deck.

12 ft

6 ft

Slide 2- 18

continued

Understand The figure can be viewed as a rectangle and a semi-circle. The diameter of the circle is the same as the width of the rectangle.

Plan We find the area of both shapes and add them together to get the total area.

A = Area of the rectangle + Area of the semi-circle.

A +lw 21

2 r

Slide 2- 19

continuedExecute Replace the variables with the corresponding values and calculate.

212+A lw r

21212 ft 6 ft 3.14 3 ftA

21212 ft 6 ft 3.14 9 ftA

2 272 ft 14.13 ftA 286.13 ftA

Slide 2- 20

continued

Answer The total area is 86.13 square feet.Check Verify the reasonableness of the answer using

estimation. Suppose the figure had been a rectangle measuring 6 ft by 15 ft. We would expect the area to be slightly larger than that of the actual figure. A = (6 ft.)(15ft.) = 90 ft2, which indicates that 86.13 square feet is reasonable.

Slide 2- 21

Sometimes, a figure may be placed within another figure and we must calculate the area of the region between the outside and inside figure.

Slide 2- 22

Example 4

A family room is being carpeted. Determine the area to be carpeted excluding the area of the semicircular hearth of the fireplace.

12 ft.

14 ft.

3 ft.

3 ft.

Slide 2- 23

continued

Understand The entire room will be carpeted excluding the shaded semi-circle.

Plan We can exclude the area of the fireplace by finding the area of the entire room and subtracting the area of the fireplace.A = Area of the rectangle – area of the semi-circle

A = lw – 2

2

r

Slide 2- 24

continued

Execute

Answer The area to be carpeted is 142.88 square feet.Check Estimate the area of the semi-circle by dividing

the room into six equal parts. The fireplace would occupy approximately one sixth of

the room. Since , the answer is reasonable.

23.14 4

14 122

A

168 25.12 A

2142.88 ftA

168 25.12 6.9

2

2

rA lw

12 ft.

14 ft.

3 ft.

3 ft.

Slide 2- 25

Nongeometric Formulas:

The distance, d, an object travels given its rate, r, and the time of travel, t: d = rt

The average rate of travel, r, given the total distance, d, and total time, t:

dr

tThe voltage, V, in a circuit with a current, i, in amperes (A), and a resistance, R, in ohms, : V = iR

The temperature in degrees Celsius given degrees Fahrenheit: 5

329

C F

The temperature in degrees Fahrenheit given degrees Celsius: 9

325

F C

Slide 2- 26

Example 5

A truck driver begins a delivery at noon and travels 120 miles before taking a 30 minute break. He then travels 136 miles, arriving at his destination at 5 pm. What was his average driving rate?

Understand We are given travel distances and times, and we are to find the driving rate.

Plan To find the average rate, we first need the total distance traveled and the total time spent driving. We can use the formula .

dr

t

Slide 2- 27

continued

Now we calculate the average rate.

Answer His average driving rate was Check We can use d = rt to verify.

56.8 4.5d

Execute Total distance = 120 + 136 = 256 miles = d

Total time driving = 5 – 0.5 = 4.5 hours = t

256 miles

4.5 hoursr

56.8 mphr

56.8 mph.

255.6 milesd

Slide 2- 28

Determine which of the following is a solution for the equation.

a) –5

b) – 1

c) 2

d) 4

6 4 2 4 3 6 x x

2.1

Determine which of the following is a solution for the equation.

a) –5

b) – 1

c) 2

d) 4

6 4 2 4 3 6 x x

2.1

A thermometer reads 30° C: what is this temperature in degrees Fahrenheit?

a) 34F

b) 49F

c) 86F

d) 112F

932

5 F C

2.1

A thermometer reads 30° C: what is this temperature in degrees Fahrenheit?

a) 34F

b) 49F

c) 86F

d) 112F

932

5 F C

2.1

The Addition Principle of Equality2.22.2

1. Determine whether a given equation is linear.2. Solve linear equations in one variable using the

addition principle.3. Solve equations with variables on both sides of

the equal sign.4. Solve identities and contradictions.5. Solve application problems.

Linear equation: An equation in which each variable term contains a single variable raised to an exponent of 1.

Linear equation in one variable: An equation that can be written in the form ax + b = c, where a, b, and c are real numbers and a 0.

Slide 2- 34

Example 1

Determine whether the equation is linear or nonlinear.a. 6y + 8 = 12Answer This equation is linear because the variable y

has an exponent of 1.

b. 8x3 + 2x = 9

Answer This equation is nonlinear because the variable in the term 8x3 has an exponent of 3.

Slide 2- 35

The Addition Principle of EqualityIf a = b, then a + c = b + c is true for all real numbers a, b, and c.

Using the Addition Principle of EqualityTo use the addition principle of equality to clear a term in an equation, add the additive inverse of that term to both sides of the equation. (That is, add or subtract appropriately so that the term you want to clear becomes 0).

Slide 2- 36

Example 2

Solve and check. x – 19 = –34

Solution

To isolate x, we need to clear –19. x – 19 = –34 + 19 + 19

x + 0 = –15 x = 15

Since we added 19 to the left side, we must add 19 to the right side as well.

Add 19 to the left-hand side so that –19 + 19 = 0.

Slide 2- 37

continued

Check Replace the x in the original equation with –15 and verify that the equation is true.

True, so –15 is the solution.

19 34 x

15 19 34 34 34

Replace x with –15 .

Slide 2- 38

Some equations have expressions that can be simplified. If like terms are on the same side of the equation, we combine the like terms before isolating the variable.

If the equation to be solved contains parentheses, we use the distributive property to clear the parentheses before isolating the variable.

Slide 2- 39

Example 3

Solve and check. 2x – 1.2 + 6.7 – x = −3.4 + 5

Solution Simplify the expressions; then isolate the x. 2x – 1.2 + 6.7 – x = −3.4 + 5

x + 5.5 = 1.6

− 5.5 = − 5.5 x + 0 = −3.9

Check Replace x in the original equation with −3.9 and verify the equation is true. We will leave this to the viewer.

x = −3.9

Slide 2- 40

Example 4

Solve and check. 6x – 7 = 5x – 19Solution Use the addition principle to get the variable

terms together on the same side of the equal sign. 6x – 7 = 5x – 19

− 5x − 5x

x – 7 = –19+ 7 + 7

x = –12

Slide 2- 41

continued

Check 6x – 7 = 5x – 19

6(−12) – 7 = 5(−12) – 19

−72 – 7 = −60 – 19

−79 = −79

True; so −12 is the solution.

Slide 2- 42

Example 5

Solve and check. Solution Simplify both sides of the equation. Then

isolate n.

4 8 6 3 8 n n n n

4 8 6 3 8 n n n n

4 8 6 18 8 n n n n

3 8 2 18 n n3 3 n n

8 18 n18 18

10 n

Distribute minus sign. Distribute the 6.

Combine like terms.

Add 3n to both sides.

Add 18 to both sides to isolate the n.

Slide 2- 43

continued

Check

True, so the solution is 10.

4 8 6 3 8 n n n n

10 4 10 8 6 10 3 8 10 10 40 8 6 10 3 80

10 48 6 7 80

10 48 42 80

38 38

Replace n in the original equation with 10 and verify that the equation is true.

Slide 2- 44

Solving Linear EquationsTo solve linear equations requiring the addition principle only:

1. Simplify both sides of the equation as needed. a. Distribute to clear parentheses. b. Combine like terms.2. Use the addition principle so that all variable terms

are on one side of the equation and all constants are on the other side. Then combine like terms.

Tip: Clear the variable term that has the lesser coefficient to avoid negative coefficients.

Slide 2- 45

In general, a linear equation in one variable has only one real-number solution. However, there are two special cases: an equation in which every real number is a solution and one that has no solution.

Identity: An equation that has every real number as a solution (excluding any numbers that cause an expression in the equation to be undefined).

Contradiction: An equation that has no real number solution

Slide 2- 46

Recognizing an IdentityWhen solving a linear equation, if after simplifying each side of the equation the expressions are identical, the equation is an identity and every real number for which the equation is defined is a solution.

Slide 2- 47

Example 6

Solve and check. 2 + 14x – 9 = 7(2x + 1) – 14 Solution Simplify both sides of the equation. Then

isolate the variable.

Check Every real number is a solution for an identity, so any number we chose will check in the original equation.

2 + 14x – 9 = 7(2x + 1) – 14

2 + 14x – 9 = 14x + 7 – 14

14x – 7 = 14x – 7

Slide 2- 48

Example 7

Solve and check. 11x + 6 = 11x + 4Solution

Check Because the variable terms, 11x, are identical on both sides of the equal sign, replacing x with any number will yield an identical product. The equation cannot be true.

−11x0 + 6 = 0 + 4

6 = 4

11x + 6 = 11x + 4−11x

Because the equation is a contradiction, it has no solution.

Slide 2- 49

Example 8

Raul wants to buy a membership to a gym which costs $395 for the year. He currently has saved $149. How much more does he need?

Understand We are given the total required for the membership and the amount he currently has. We must find how much more he needs.

Plan Let x represent the amount Raul needs. Write the equation, then solve.

Slide 2- 50

continued

Execute Current amount + amount needed = 395149 + x = 395

Answer Raul needs $246 to buy the membership. Does $149 plus the additional $246 equal

$395? 149 + 246 = 395

395 = 395

149 + x = 395–149 –149

0 + x = 246

x = 246

Subtract 149 from both sides to isolate x.

Slide 2- 51

Solve for t. 8t + 5 –14t = 5t + 3 – 9t

a) t = 5

b) t = 3

c) t = 1

d) t = –1

2.2

Solve for t. 8t + 5 –14t = 5t + 3 – 9t

a) t = 5

b) t = 3

c) t = 1

d) t = –1

2.2

An 8 ft board is cut into 3 pieces. If one is 3 ½ ft and the second is 2 ¼ ft, what is the length of the third piece?

a) 1 ¾ ft

b) 2 ¼ ft

c) 2 ½ ft

d) 3 ¼ ft

2.2

An 8 ft board is cut into 3 pieces. If one is 3 ½ ft and the second is 2 ¼ ft, what is the length of the third piece?

a) 1 ¾ ft

b) 2 ¼ ft

c) 2 ½ ft

d) 3 ¼ ft

2.2

The Multiplication Principle of Equality2.32.3

1. Solve linear equations using the multiplication principle.

2. Solve linear equations using both the addition and the multiplication principles.

3. Use the multiplication principle to clear fractions and decimals from equations.

4. Solve application problems.

The Multiplication Principle of EqualityIf a = b, then ac = bc is true for all real numbers a, b, and c, where c 0.

Using the Multiplication Principle of EqualityTo use the multiplication principle of equality to clear a coefficient in an equation, multiply both sides of the equation by the multiplicative inverse of that coefficient or divide both sides by the coefficient.

Slide 2- 57

Example 1

Solve and check.

Solution

5 3

4 8 x

5 3

4 8 x

4 45

5 5

3

4 8x

1

11

1

Clear the coefficient by multiplying both sides by its multiplicative inverse, .4

5

5

4

3

10x

2

1

Slide 2- 58

continued

Check

True, therefore is correct.

5 3

4 8 x

3

10

5 3

4 8

3 3

8 8

Replace x in the original equation with and verify that the equation is true.

3

10

3

10

1

2

Slide 2- 59

Solving Linear EquationsTo solve linear equations in one variable,

1. Simplify both sides of the equation as needed. a. Distribute to clear parentheses. b. Combine like terms.2. Use the addition principle so that all variable terms are

on one side of the equation and all constants are on the other side. (Clear the variable term with the lesser coefficient to avoid negative coefficients.) Then combine like terms.

3. Use the multiplication principle to clear any remaining coefficient.

Slide 2- 60

Example 2

Solve and check. Solution

8 9 7x

9 9

= 8 8

2x

The answer checks so x = −2.

8 9 7x

8 0 16x 8 16x

Slide 2- 61

Example 3

Solve and check. Solution

7 11 2 46y y

2 2 y y

= 5 5

7 11 2 46y y

5 11 0 46y

11 1 1

5 35y

5 11 46y

7y The answer checks so y = −7.

Slide 2- 62

Example 4

Solve and check. Solution

2 5 5 3 2 1 y y

2 5 5 3 2 1 y y

2 5 25 3 6 1 y y

5 23 3 7 y y5 5 y y

0 23 8 7 y

Distribute to clear parentheses.

Combine like terms.

Add 5y to both sides (–5y has the lesser coefficient).

Slide 2- 63

continued

Check

0 23 8 7 y7 7 16 8 0 y16 8

8 8

y

2 y

Add 7 to both sides.

Divide both sides by 8 to clear the 8 coefficient.

2 5 5 3 2 1 y y

25 3 2 122 5

2 5 3 3 4 1 2 15 12 1

13 13 True

Replace y in the original equation with –2 and verify that the equation is true.

Therefore 2 is correct.

Slide 2- 64

If the equation contains fractions, we multiply both sides by a number that will clear all the denominators. We could multiply both sides by any multiple of the denominators; however, using the LCD (least common denominator) results in the simplest equations.

The multiplication principle can also be used to clear decimals in an equation. We can clear decimals by multiplying both sides of the equation by an appropriate power of 10. The power of 10 we use depends on the decimal number with the most decimal places.

Slide 2- 65

Example 5

Solve and check.Solution

7 1 3 1

8 4 4 16x x x

7 1 3 1

8 4 4 1616 16x x x

16 16 16 1616

1 1 1

7 1 3 1

4 4 1618x x x

14 4 12 1 16x x x 26 4 1 16x x

10 5x 1

2x

42

1 1 1

4 1

1

Slide 2- 66

Example 6

Solve and check.Solution

16.3 7.2 26.9 n

16.3 7.2 2 .1 1090 6 n

16.3 7.2 26.9 n

163 72 269 n163 163

0 72 432 n

72 432

72 72

n

6n

16.3 7.2 66 2 .9

16.3 7.2 26.9 n

16.3 43.2 26.9

26.9 26.9

Check

Slide 2- 67

Solving Linear EquationsTo solve linear equations in one variable,

1. Simplify both sides of the equation as needed. a. Distribute to clear parentheses. b. Clear fractions or decimals by multiplying through

by the LCD. In the case of decimals, the LCD is the power of 10 with the same number of zero digits as decimal places in the number with the most decimal places. (Clearing fractions and decimals is optional.)c. Combine like terms

Slide 2- 68

continued Solving Linear Equations

2. Use the addition principle so that all variable terms are on one side of the equation and all constants are on the other side. (Clear the variable term with the lesser coefficient.) Then combine like terms.

3. Use the multiplication principle to clear any remaining coefficient.

Slide 2- 69

Example 7

The perimeter of the figure shown is 72 inches. Find the width and the length.

x +11

x

Understand The width is represented by x and the length is represented by x + 11. We are given the perimeter, so we can use the formula P = 2l + 2w.

Slide 2- 70

continued

Plan In the perimeter formula, replace P with 72, l with x + 11, w with x, and solve for x.

Execute 2 2 P l w

12 2 217 x x

72 2 22 2 x x

72 4 22 x

22 22 50 4 0 x

Distribute.

Combine like terms.

Subtract 22 from both sides.

50 4

4 4

xDivide both sides by 4.

12.5 xSlide 2- 71

continued

Answer The width is 12.5 inches. To find the length, we evaluate the expression that represents the

length, x + 11, with x = 12.5 inches Length = 12.5 + 11 = 23.5 inches

Check

It checks.

2 2

2 23.5 2 12.5

47 25

72

P l w

P

P

P

Slide 2- 72

Solve. 8m + 6 = 3(12 + 2m)

a) 3

b) 5

c) 8

d) 15

2.3

Solve. 8m + 6 = 3(12 + 2m)

a) 3

b) 5

c) 8

d) 15

2.3

Solve. 0.8 – 4(a – 1) = 0.2 + 3(4 – a)

a) 10

b) 7.4

c) 8.6

d) 20.6

2.3

Solve. 0.8 – 4(a – 1) = 0.2 + 3(4 – a)

a) 10

b) 7.4

c) 8.6

d) 20.6

2.3

Applying the Principles to Formulas2.42.4

1. Isolate a variable in a formula using the addition and multiplication principles.

Isolating a Variable in a FormulaTo isolate a particular variable in a formula, treat all other variables like constants and isolate the desired variable using the outline for solving equations.

Slide 2- 78

Example 1

Isolate W in the formula Z = W – Y.

Solution Z = W – Y

+ Y + Y Z + Y = W + 0

Z + Y = W

To isolate W we must clear Y. Because Y is subtracted from W, we add Y to both sides.

Slide 2- 79

Example 2

Isolate l in the formula for the volume of a box, V = lwh.

SolutionV = lwh

To isolate l we must clear w and h. Because w and h are multiplying l, we divide both sides by w and h.

V l

wh

wh

wh

Vl

wh

Slide 2- 80

Example 3

Isolate m in the formula jm + c = n.

Solutionjm + c = n

c c jm = n – c

Isolate jm by subtracting c from both sides.

Isolate m by dividing both sides by j.

jm

j

n c

jn c

mj

Slide 2- 81

Isolate a:

a)

b)

c)

d)

WN

a

Wa

N

Na

W

a NW

a N W

2.4

Isolate a:

a)

b)

c)

d)

WN

a

Wa

N

Na

W

a NW

a N W

2.4

Isolate a:

a)

b)

c)

d)

180a b c

180a b c

180a b c

180a

b c

180 ca

b

2.4

Isolate a:

a)

b)

c)

d)

180a b c

180a b c

180a b c

180a

b c

180 ca

b

2.4

Translating Word Sentences to Equations2.52.5

1. Translate sentences to equations using key words, then solve.

Key Words and Their TranslationsAddition Translation Subtraction Translation

The sum of x and 3

x + 3 The difference of x and 3

x – 3

h plus k h + k h minus k h – k

7 added to t t + 7 7 subtracted from t t – 7

3 more than a number

n + 3 3 less than a number

n – 3

y increased by 2 y + 2 y decreased by 2 y – 2

Note: Since addition is a commutative operation, it does not matter in what order we write the translation.

Note: Subtraction is not a commutative operation; therefore, the way we write the translation matters.

Slide 2- 87

Key Words and Their TranslationsMultiplication Translation Division Translation

The product of x and 3

3x The quotient of x and 3

x 3, x/3

h times k hk h divided by k h k, h/k

Twice a number 2n h divided into k k h, k/h

Triple the number

3n The ratio of a to b a b, a/b

Two-thirds of a number

Note: Like addition, multiplication is a commutative operation, it does not matter in what order we write the translation.

Note: Division is like subtraction in that it is not a commutative operation; therefore, the way we write the translation matters.

2

3n

Slide 2- 88

Key words for an equal sign:is equal to is yields

is the same as produces results in

Translating Word SentencesTo translate a word sentence to an equation, identify the unknown(s), constants, and key words; then write the corresponding symbolic form.

Slide 2- 89

Example 1

The sum of thirty-five and a number is equal to eighteen. Translate to an equation, then solve for the number.

Understand The key word sum indicates addition,is equal to indicates an equal sign, and a number indicates a variable.

Plan Translate the key words to an equation, and then solve the equation. We will use n as the variable.

Slide 2- 90

continued

Execute Translate:

The sum of thirty-five and a number is equal to 18.

Solve: 35 + n = 1835 35 0 + n = 17 n = 17

35 + n = 18

Answer

Check

35 + n = 18 35 + (–17) = 18 18 = 18

Slide 2- 91

Example 2Two-thirds of a number is negative seven-eighths. Translate to an equation and then solve.

Understand When of is preceded by a fraction, it means multiply. The word is means an equal sign.

Plan Translate the key words and then solve.Execute Translate:Two-thirds of a number is negative seven-eighths.

2

3 n 7

8

Slide 2- 92

continued

Solve: 2 7

3 8n

3 2 7 3

2 3 8 2n

Clear the coefficient 2/3 by multiplying both sides by its reciprocal 3/2.

Answer21

16n

Check 2 7

3 8n

2 21 7

3 16 8

7 7

8 8

1

81

7

Slide 2- 93

Example 3Eight less than five times a number is equal to thirty-seven. Translate to an equation, and then solve. Understand Less than indicates subtraction in reverse order,times indicates multiplication,is equal to indicates an equal sign.

Plan Translate to an equation using the key words and then solve the equation.

Slide 2- 94

continuedEight less than five times a number is equal to thirty-seven.

Solve: 5n – 8 = 37 + 8 + 85n + 0 = 45 5n = 45 5 5 n = 9

85n = 37

Answer

Check:

5n – 8 = 375(9) – 8 = 37 45 – 8 = 37 37 = 37

Slide 2- 95

Example 4Nine times the sum of a number and seven subtracted from three times the number results in negative twenty-seven.Understand Times means multiply,subtracted from indicates subtraction,sum means addition,result in indicates an equal sign.

Plan Translate the key words to an equation and then solve.

Slide 2- 96

continuedNine times the sum of a number and seven subtracted from three times the number results in negative twenty-seven.Execute: 3x 9(x + 7) = 27

3x – 9x – 63 = 27 6x – 63 = 27 + 63 +63 6x + 0 = 36 6x = 36

6 6 x = 6

Distribute to clear the parentheses.

Simplify.

Add 63 to both sides.

Divide both sides by 6.

Answer

Check:

3x – 9(x + 7) = 273(6) – 9(6 + 7) = 27 18 – 9(1) = 27

27 = 27

Slide 2- 97

Translate: Thirty-two less than three times a number is fifteen.

a) 32 – 3x = 15

b) 3x – 32 = 15

c) 32 + 3x = 15

d) 3x 32 = 15

2.5

Translate: Thirty-two less than three times a number is fifteen.

a) 32 – 3x = 15

b) 3x – 32 = 15

c) 32 + 3x = 15

d) 3x 32 = 15

2.5

Translate: The quotient of four less than a number and five is the same as the number divided by eight.

a)

b)

c)

d) 4

5 8

n n

48

5

nn

5( 4) 8n n

5 8

4n n

2.5

Translate: The quotient of four less than a number and five is the same as the number divided by eight.

a)

b)

c)

d) 4

5 8

n n

48

5

nn

5( 4) 8n n

5 8

4n n

2.5

Solving Linear Inequalities2.62.6

1. Represent solutions to inequalities graphically and using set notation.

2. Solve linear inequalities.3. Solve problems involving linear inequalities.

Not all problems translate to equations. Sometimes a problem can have a range of values as solutions.In mathematics we write inequalities to describe situations where a range of solutions is possible.

< is less than

> is greater than

is less than or equal to

is greater than or equal to

Slide 2- 103

Linear inequality: An inequality containing expressions in which each variable term contains a single variable with an exponent of 1.

Examples of linear inequalities:

x > 5 n + 2 < 6 2(y – 3) 5y – 9

Set builder notation

{ x | x 5}

The set of all x such that x is greater than or equal to 5.

Slide 2- 104

We can graph solution sets for inequalities on a number line.Since the solution set for x 5 contains 5 and every real number to the right of 5, we draw a dot (or solid circle) at 5 and shade to the right of 5.

x < 2: Open circle on 2 and shade to the left of 2

Parentheses and brackets can also be used on graphs instead of open and solid circles.

)

[

Slide 2- 105

Graphing Inequalities To graph an inequality on a number line,1. If the symbol is or , draw a bracket (or solid

circle) on the number line at the indicated number open to the left for and to the right for . If the symbol is < or >, draw a parenthesis (or open circle) on the number line at the indicated number open to the left for < and to the right for >.

2. If the variable is greater than the indicated number, shade to the right of the indicated number. If the variable is less than the indicated number, shade to the left of the indicated number.

Slide 2- 106

Example 1

Write the solution set in set-builder notation and interval notation, then graph the solution set.

a. x 2 b. n > 3Solutiona. x 2

Set-builder notation: {x|x 2}Interval notation: (, 2]Graph: ]

Slide 2- 107

continued

b. n > 3

Set-builder notation: {n|n > 3}Interval notation: (3, )

Graph: (

Slide 2- 108

Inequalities containing two inequality symbols are called compound inequalities. Compound inequalities are useful in writing a range of values between two numbers.For example, 1 < x < 6x can be any number greater than 1 and less than 6. The solution set contains every real number between 1 and 6, but not 1 and 6.

Set-builder notation: {x|1 < x < 6}Interval notation: (1, 6)Graph: ( )

Slide 2- 109

Example 2

Write the solution set for 7 < x 3 in set-builder notation and interval notation, then graph the solution set.Solution

Set-builder notation: {x|7 < x 3}Interval notation: (7, 3]Graph: ( ]

Slide 2- 110

To solve inequalities, we will follow essentially the same process as for solving equations.

The Addition Principle of Inequality

If a < b, then a + c < b + c is true for all real numbers a, b, and c. The principle also holds true when < is replaced with >, , or .

Slide 2- 111

Example 3

Solve x + 8 7 and write the solution set in set-builder notation and interval notation; then graph the solution set.Solution

Set-builder notation: {x| x −1}Interval notation: (∞, −1]Graph:

]

x + 8 7−8 −8x −1

Slide 2- 112

The multiplication principle, on the other hand, does not work as neatly as it did for equations.

The Multiplication Principle of Inequality

If a and b are real numbers, where a < b, then ac < bc is true if c is a positive real number.

If a and b are real numbers, where a < b, then ac > bc is true if c is a negative real number. The principle also holds true for >, , and .

Slide 2- 113

Example 4Solve and write the solution set in set-builder notation and interval notation. Then graph the solution set.

Solution

Set-builder notation: {x|x 4}Interval notation: (, 4]Graph:

7 28x 7 28x 7 28

7 7

x

4x

Because we divided both sides by a negative number, we reversed the direction of the inequality symbol.

]

Slide 2- 114

Solving Linear Inequalities To solve linear inequalities,1. Simplify both sides of the inequality as needed.

a. Distribute to clear parentheses.b. Clear fractions or decimals by multiplying through by the LCD just as we did for equations. (Clearing fractions and decimals is optional.)c. Combine like terms.

2. Use the addition principle so that all variable terms are on one side of the inequality and all constants are on the other side. Then combine like terms.

3. Use the multiplication principle to clear any remaining coefficient. If you multiply (or divide) both sides by a negative number, reverse the direction of the inequality symbol.

Slide 2- 115

Example 5Solve 8x + 13 > 3x – 12.Solution 8x + 13 > 3x – 12

3x 3x 5x + 13 > 0 12 5x + 13 > 12

13 13 5x + 0 > 25 5x > 25 5 5

x > 5

Subtract 3x from both sides.

Subtract 13 from both sides.

Divide both sides by 5 to isolate x.

Set builder notation: {x|x > 5}Interval notation: (5, )Graph: (

Slide 2- 116

Key Words and Their TranslationsLess Than: Greater Than:

A number is less than seven.

n < 7 A number is greater than two.

n > 2

A number must be smaller than five.

n < 5 A number must be greater than three.

n > 3

A number must be more than negative six.

n > 6

Less Than or Equal to:

Greater Than or Equal to:

A number is at most nine.

n 9 A number is at least two.

n 2

The maximum is fourteen.

n 14 The minimum is eighteen.

n 18

Slide 2- 117

Example 6Seven-eighths of a number is at least twenty-one. Translate to an inequality, then solve.Understand Because the word of is preceded by a fraction, it means multiplication. The key words at least indicate a greater-than or equal-to symbol.Plan Translate the key words, then solve. Use n for the variable.Execute Translate. Seven-eighths of a number is at least twenty-one.

7

18

2 n

Slide 2- 118

continued

Solve: 7

218

n

8 7 8 21

7 8 7 1n Multiply both sides by 8/7 to isolate n.

3

11

11

1

24n Answer

Check Verify that 24 and any number greater than 24 satisfy the original sentence.

Slide 2- 119

continued

Test 24: Test a number greater than 24:

Is of 24 at least 21?

7

87 24

218 1

21 21

Is of 32 at least 21?7

87 32

218 1

28 21

True True

3

1

4

1

Slide 2- 120

Example 7A park board is planning to enclose a new playground area with fence. Cost restricts the board to a total of 320 feet of fencing materials. The board wants the playground to be 75 feet wide. What is the maximum length of the playground?

Understand The fence surrounds the playground, so that 320 feet is the maximum perimeter of the playground. The width is to be 75 feet, we need to find the length.

Slide 2- 121

continuedPlan Formula for perimeter of a rectangle: P = 2l + 2w

320 is the maximum perimeter so we write an inequality so the expression used to calculate perimeter is less than or equal to 320. Because the width is to be 75 feet we replace w with 75.

Execute Translate:The perimeter must be less than or equal to 320.

2l + 2(75) 320

Slide 2- 122

continued

Solve: 2l + 2(75) 3202l + 150 320 Subtract 150 from both

sides.

150 150 2l 170 Divide both sides by 2.

2 2 l 85

Answer The length must be less than or equal to 85 feet, which means the maximum length is 85 feet.

Slide 2- 123

Solve 9x + 12 > 3x – 18.

a) x > 5

b) x > 5

c) x < 5

d) x < 1

2.6

Solve 9x + 12 > 3x – 18.

a) x > 5

b) x > 5

c) x < 5

d) x < 1

2.6

Solve. Eight times a number less thirty-two is at least seventy-two.

a) x 5

b) x 5

c) x 13

d) x 13

2.6

Solve. Eight times a number less thirty-two is at least seventy-two.

a) x 5

b) x 5

c) x 13

d) x 13

2.6