chapter 1: equations and inequalities big ideas: 1.use properties to evaluate and simplify...
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CHAPTER 1:EQUATIONS AND
INEQUALITIES
BIG IDEAS:1. Use properties to evaluate and simplify
expressions2. Use problem solving strategies and verbal
models3. Solve linear and absolute value equations
and inequalities
What is the difference between a daily low
temperature of -5 and a daily high
temperature of 18?
LESSON 1: APPLY PROPERTIES OF
REAL
ESSENTIAL QUESTION
How are addition and subtraction related and how are multiplication and division related?
• Opposite: additive inverse: the opposite of b is –b
• Reciprocal: multiplicative inverse: the reciprocal of a = 1/a.
VOCABULARY
Properties
EXAMPLE 1 Graph real numbers on a number line
Graph the real numbers – and 3 on a number line.5
4
SOLUTION
Note that – = –1.25. Use a calculator to approximate 3 to the nearest tenth:
5
4
3 1.7. (The symbol means is approximately equal to.)
So, graph – between –2 and –1, and graph 3 between
1 and 2, as shown on the number line below.
5
4
EXAMPLE 3 Identify properties of real numbers
Identify the property that the statement illustrates.
a. 7 + 4 = 4 + 7
b. 13 = 11
13
SOLUTION
Inverse property of multiplication
Commutative property of addition
SOLUTION
EXAMPLE 4 Use properties and definitions of operations
Use properties and definitions of operations to show that a + (2 – a) = 2. Justify each step.
SOLUTION
a + (2 – a) = a + [2 + (– a)] Definition of subtraction
= a + [(– a) + 2] Commutative property of addition
= [a + (– a)] + 2 Associative property of addition
= 0 + 2 Inverse property of addition
= 2 Identity property of addition
Identify the property that the statement illustrates.
4. 15 + 0 = 15
SOLUTION
Identity property of addition.
Associative property of multiplication.
SOLUTION
3. (2 3) 9 = 2 (3 9)
GUIDED PRACTICE for Examples 3 and 4
Identify the property that the statement illustrates.
5. 4(5 + 25) = 4(5) + 4(25)
SOLUTION
Identity property of multiplication.
Distributive property.
SOLUTION
6. 1 500 = 500
GUIDED PRACTICE for Examples 3 and 4
Use properties and definitions of operations to show that the statement is true. Justify each step.
SOLUTION
Def. of division
GUIDED PRACTICE for Examples 3 and 4
1
b= b ( 4) Comm. prop. of multiplication
Assoc. prop. of multiplication1
b= (b ) 4
= 1 4 Inverse prop. of multiplication
Identity prop. of multiplication= 4
1
b= b (4 )b (4 b)
7. b (4 b) = 4 when b = 0
Use properties and definitions of operations to show that the statement is true. Justify each step.
SOLUTION
8. 3x + (6 + 4x) = 7x + 6
GUIDED PRACTICE for Examples 3 and 4
Assoc. prop. of addition
Combine like terms.
Comm. prop. of addition3x + (6 + 4x) = 3x + (4x + 6)
= (3x + 4x) + 6
= 7x + 6
ESSENTIAL QUESTION
How are addition and subtraction related and how
are multiplication and division related?
They are inverses of one another. Subtraction is defined as adding the
opposite of the number being subtracted. Division is defined as multiplying by the
reciprocal of the divisor.
Jill has enough money for a total
of 32 table decorations and
wall decorations. If n is the
number of table decorations,
write an expression for the number of
wall decorations she can buy.
LESSON 2: EVALUATE AND
SIMPLIFY ALGEBRAIC
EXPRESSION
ESSENTIAL QUESTION
When an expression involves more than one operation, in what order
do you do the operations?
• Power: an expression formed by repeated multiplication of the same factor
• Variable: a letter that is used to represent one or more numbers
• Term: each part of an expression separated by + and – signs
• Coefficient: the number that leads a variable
• Identity: a statement that equates to two equivalent expressions
VOCABULARY
EXAMPLE 1 Evaluate powers
a. (–5)4
b. –54
= (–5) (–5) (–5) (–5) = 625
= –(5 5 5 5) = –625
EXAMPLE 2 Evaluate an algebraic expression
Evaluate –4x2 – 6x + 11 when x = –3.
–4x2 – 6x + 11 = –4(–3)2 – 6(–3) + 11 Substitute –3 for x.
= –4(9) – 6(–3) + 11 Evaluate power.
= –36 + 18 + 11 Multiply.
= –7 Add.
EXAMPLE 3 Use a verbal model to solve a problem
Craft Fair
You are selling homemade candles at a craft fair for $3 each. You spend $120 to rent the booth and buy materials for the candles.
• Write an expression that shows your profit from
selling c candles.
• Find your profit if you sell 75 candles.
EXAMPLE 3 Use a verbal model to solve a problem
SOLUTION
STEP 1 Write: a verbal model. Then write an algebraic expression. Use the fact that profit is the difference between income and expenses.
–
An expression that shows your profit is 3c – 120.
3 c – 120
EXAMPLE 3 Use a verbal model to solve a problem
STEP 2 Evaluate: the expression in Step 1 when c = 75.
3c – 120 = 3(75) – 120 Substitute 75 for c.
= 225 – 120
= 105 Subtract.
ANSWER Your profit is $105.
Multiply.
GUIDED PRACTICE for Examples 1, 2, and 3
Evaluate the expression.
631.
–262.
SOLUTION
216
SOLUTION
–64
GUIDED PRACTICE for Examples 1, 2, and 3
(–2)63.
5x(x – 2) when x = 64.
SOLUTION
64
SOLUTION
120
GUIDED PRACTICE for Examples 1, 2, and 3
3y2 – 4y when y = – 25.
SOLUTION
20
(z + 3)3 when z = 16.
SOLUTION
64
GUIDED PRACTICE for Examples 1, 2, and 3
What If? In Example 3, find your profit if you sell 135 candles.7.
ANSWER Your profit is $285.
EXAMPLE 4 Simplify by combining like terms
a. 8x + 3x = (8 + 3)x Distributive property
= 11x Add coefficients.
b. 5p2 + p – 2p2 = (5p2 – 2p2) + p Group like terms.
= 3p2 + p Combine like terms.
c. 3(y + 2) – 4(y – 7) = 3y + 6 – 4y + 28 Distributive property
= (3y – 4y) + (6 + 28) Group like terms.
= –y + 34 Combine like terms.
EXAMPLE 4 Simplify by combining like terms
d. 2x – 3y – 9x + y = (2x – 9x) + (– 3y + y) Group like terms.
= –7x – 2y Combine like terms.
GUIDED PRACTICE for Example 5
8. Identify the terms, coefficients, like terms, and constant terms in the expression 2 + 5x – 6x2 + 7x – 3. Then simplify the expression.
SOLUTION
Terms:
Coefficients:
Like terms:
Constants:
–6x2 +12x – 1Simplify:
2, 5x, –6x2 , 7x, –3
5 from 5x, –6 from –6x2 , 7 from 7x
5x and 7x, 2 and –3
2 and –3
GUIDED PRACTICE for Example 5
9. 15m – 9m
SOLUTION
6m
Simplify the expression.
10. 2n – 1 + 6n + 5
SOLUTION
8n + 4
GUIDED PRACTICE for Example 5
11. 3p3 + 5p2 – p3
SOLUTION
2p3 + 5p2
12. 2q2 + q – 7q – 5q2
SOLUTION
–3q2 – 6q
GUIDED PRACTICE for Example 5
13. 8(x – 3) – 2(x + 6)
SOLUTION
6x – 36
14. –4y – x + 10x + y
SOLUTION
9x –3y
ESSENTIAL QUESTION
When an expression involves more than one operation, in what order
do you do the operations?Order of operations:
Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction
On a blank sheet of paper,
complete #1-13 ODD on P16 in the blue Quiz section.
Please turn into the
homework bin when
finished.
LESSON 3: SOLVE LINEAR EQUATIONS
ESSENTIAL QUESTION
What are the steps for solving a linear equation?
• Equation: a statement that two expressions are equal
• Linear equation: may be written in the form ax + b = 0; no exponents
• Solution: a number that makes a true statement when substituted into an equation
• Equivalent equations: two equations that have the same solutions
VOCABULARY
EXAMPLE 1 Solve an equation with a variable on one side
Solve 4
5x + 8 = 20.
4
5x + 8 = 20
4
5x = 12
x = (12)5
4
x = 15
Write original equation.
Subtract 8 from each side.
Multiply each side by , the reciprocal of .
544
5Simplify.
ANSWER The solution is 15.
CHECK x = 15 in the original equation.
4
54
5x + 8 = (15) + 8 = 12 + 8 = 20
EXAMPLE 2 Write and use a linear equation
During one shift, a waiter earns wages of $30 and gets an additional 15% in tips on customers’ food bills. The waiter earns $105. What is the total of the customers’ food bills?
Restaurant
SOLUTION
Write a verbal model. Then write an equation. Write 15% as a decimal.
EXAMPLE 2 Write and use a linear equation
105 = 30 + 0.15x
75 = 0.15x
500 = x
Write equation.
Subtract 30 from each side.
Divide each side by 0.15.
The total of the customers’ food bills is $500.
ANSWER
GUIDED PRACTICE for Examples 1 and 2
Solve the equation. Check your solution.
1. 4x + 9 = 21
ANSWER The solution is x = 3.
2. 7x – 41 = – 13
ANSWER The solution is x = 4.
ANSWER The solution is 5.
3. 35
– x + 1 = 4
GUIDED PRACTICE for Examples 1 and 2
A real estate agent’s base salary is $22,000 per year. The agent earns a 4% commission on total sales. How much must the agent sell to earn $60,000 in one year?
4. REAL ESTATE
The agent must sell $950,000 in a year to each $ 60000
ANSWER
EXAMPLE 3 Standardized Test Practice
SOLUTION
7p + 13 = 9p – 5
13 = 2p – 5
18 = 2p
9 = p
Write original equation.
Subtract 7p from each side.
Add 5 to each side.
Divide each side by 2.
ANSWER The correct answer is D
EXAMPLE 3 Standardized Test Practice
CHECK
7p + 13 = 9p – 5
7(9) + 13 9(9) – 5=?
63 + 13 81 – 5=?
76 = 76
Write original equation.
Substitute 9 for p.
Multiply.
Solution checks.
EXAMPLE 4 Solve an equation using the distributive property
Solve 3(5x – 8) = –2(–x + 7) – 12x.
3(5x – 8) = –2(–x + 7) – 12x
15x – 24 = 2x – 14 – 12x
15x – 24 = – 10x – 14
25x – 24 = –14
25x = 10
x =25
Write original equation.
Distributive property
Combine like terms.
Add 10x to each side.
Add 24 to each side.
Divide each side by 25 and simplify.
ANSWER The solution 25
EXAMPLE 4 Solve an equation using the distributive property
CHECK
25
3(5 – 8) –2(– + 7) – 12 25
25
=?
3(–6) –14 – 45
=?245
– 18 = – 18
25
Substitute for x.
Simplify.
Solution checks.
EXAMPLE 5 Solve a work problem
Car Wash
It takes you 8 minutes to wash a car and it takes a friend 6 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?
SOLUTION
STEP 1 Write a verbal model. Then write an equation.
EXAMPLE 5 Solve a work problem
Solve the equation for t.STEP 2
1
8t + t = 7
1
6
24( t + t) = 24 (7)1
81
6
3t + 4t = 168
7t = 168
t = 24
Write equation.
Multiply each side by the LCD, 24.
Distributive property
Combine like terms.
Divide each side by 7.
ANSWER It will take 24 minutes to wash 7 cars if you work together.
EXAMPLE 5 Solve a work problem
CHECK
You wash 24 = 3 cars and your friend washes 24 = 4 cars in 24 minutes. Together, you wash 7 cars.1
6
1
8
GUIDED PRACTICE for Examples 3, 4, and 5
Solve the equation. Check your solution.
5. –2x + 9 = 2x – 7
ANSWER The correct answer is 4.
6. 10 – x = –6x + 15
ANSWER The correct answer is 1.
7. 3(x + 2) = 5(x + 4)
ANSWER The solution is –7.
GUIDED PRACTICE for Examples 3, 4, and 5
Solve the equation. Check your solution.
8. –4(2x + 5) = 2(–x – 9) – 4x
ANSWER The solution x = – 1
x + x = 391
42
59.
ANSWER The correct answer is 60
GUIDED PRACTICE for Examples 3, 4, and 5
Solve the equation. Check your solution.
10. x + = x – 1
2
2
3
5
6
ANSWER The correct answer is 4
What If? In Example 5, suppose it takes you 9 minutes to wash a car and it takes your friend 12 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?
11.
ANSWER It will take 36 minutes to wash 7 cars if you work together.
ESSENTIAL QUESTION
What are the steps for solving a linear equation?
If the equation involves an expression in parenthesis,
remove the parentheses by using the distributive property.
Then use the properties of equality to obtain equivalent equations in a series of steps
until you obtain an equation of the form x = a.
Use the distributive property to
rewrite xy-5y as a product.
LESSON 4: REWRITE
FORMULAS AND EQUATIONS
ESSENTIAL QUESTION
What are formulas, and how are formulas used?
• Formula: an equation that relates two or more quantities, usually represented by variables
• Solve for a variable: to rewrite an equation as an equivalent equation in which the variable is on one side and does not appear on the other side; isolate the variable
VOCABULARY
EXAMPLE 1 Rewrite a formula with two variables
Solve the formula C = 2πr for r. Then find the radius of a circle with a circumference of 44 inches.
SOLUTION
C = 2πr
C2π
= r
STEP 1 Solve the formula for r.
STEP 2 Substitute the given value into the rewritten formula.
Write circumference formula.
Divide each side by 2π.
r =C
2π=
442π
7 Substitute 44 for C and simplify.
The radius of the circle is about 7 inches.ANSWER
GUIDED PRACTICE for Example 1
Find the radius of a circle with a circumference of 25 feet.1.
The radius of the circle is about 4 feet.ANSWER
GUIDED PRACTICE for Example 1
The formula for the distance d between opposite vertices of a regular hexagon is d = where a is the distance between opposite sides. Solve the formula for a. Then find a when d = 10 centimeters.
2.2a3
SOLUTION
d 3a =
2
35When d = 10cm, a = or 8.7cm
EXAMPLE 2 Rewrite a formula with three variables
SOLUTION
Solve the formula for w.STEP 1
P = 2l + 2w
P – 2l = 2w
P – 2l2
= w
Write perimeter formula.
Subtract 2l from each side.
Divide each side by 2.
Solve the formula P = 2l + w for w. Then find the width of a rectangle with a length of 12 meters and a perimeter of 41 meters.
EXAMPLE 2 Rewrite a formula with three variables
41 – 2(12)
2w =
w = 8.5
Substitute 41 for P and 12 for l.
Simplify.
The width of the rectangle is 8.5 meters.
ANSWER
Substitute the given values into the rewritten formula.
STEP 2
GUIDED PRACTICE for Example 2
Solve the formula P = 2l + 2w for l. Then find the length of a rectangle with a width of 7 inches and a perimeter of 30 inches.
3.
Length of rectangle is 8 in.ANSWER
Solve the formula A = lw for w. Then find the width of a rectangle with a length of 16 meters and an area of 40 square meters.
4.
Write of rectangle is 2.5 mw = A
lANSWER
GUIDED PRACTICE for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A =1
2bh5.
Find h if b = 12 m
and A = 84 m2.
= h2A
b
ANSWER
GUIDED PRACTICE for Example 2
Find the value of h if b = 12m and A = 84m2.
Find h if b = 12 m
and A = 84 m2.
= h2A
b
h = 14m
ANSWER
GUIDED PRACTICE for Example 2
Find b if h = 3 cm
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A =1
2bh6.
and A = 9 cm2.
= b2A
h
ANSWER
GUIDED PRACTICE for Example 2
Find b if h = 3 cm
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A =1
2bh6.
and A = 9 cm2.
b = 6cm
ANSWER
GUIDED PRACTICE for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A =1
27. (b1 + b2)h Find h if b1 = 6 in.,
b2 = 8 in., and A = 70 in.2
h =2A
(b1 + b2)
ANSWER
GUIDED PRACTICE for Example 2
Solve the formula for the variable in red. Then use the given information to find the value of the variable.
A =1
27. (b1 + b2)h Find h if b1 = 6 in.,
b2 = 8 in., and A = 70 in.2
h = 10 in.
ANSWER
EXAMPLE 3 Rewrite a linear equation
Solve 9x – 4y = 7 for y. Then find the value of y when x = –5.
SOLUTION
Solve the equation for y.STEP 1
9x – 4y = 7
–4y = 7 – 9x
y =9
4
7
4– + x
Write original equation.
Subtract 9x from each side.
Divide each side by –4.
EXAMPLE 3 Rewrite a linear equation
Substitute the given value into the rewritten equation.STEP 2
y =9
4
7
4– + (–5)
y =45
4
7
4– –
y = –13
CHECK
9x – 4y = 7
9(–5) – 4(–13) 7=?
7 = 7
Substitute –5 for x.
Multiply.
Simplify.
Write original equation.
Substitute –5 for x and –13 for y.
Solution checks.
EXAMPLE 4 Rewrite a nonlinear equation
Solve 2y + xy = 6 for y. Then find the value of y when x = –3.
SOLUTION
Solve the equation for y.STEP 1
2y + x y = 6
(2+ x) y = 6
y =6
2 + x
Write original equation.
Distributive property
Divide each side by (2 + x).
EXAMPLE 4 Rewrite a nonlinear equation
Substitute the given value into the rewritten equation.STEP 2
y =6
2 + (–3)
y = – 6
Substitute –3 for x.
Simplify.
GUIDED PRACTICE for Examples 3 and 4
Solve the equation for y. Then find the value of y when x = 2.
8. y – 6x = 7
y = 7 + 6x
y = 19
ANSWER
9. 5y – x = 13
y =x5
135
+
y = 5
ANSWER
10. 3x + 2y = 12
y = –3x2
+ 6
ANSWER
y = 3
GUIDED PRACTICE for Examples 3 and 4
Solve the equation for y. Then find the value of y when x = 2.
11. 2x + 5y = –1 12. 3 = 2xy – x 13. 4y – xy = 28
y = 14
284 – x
y =
ANSWER
y = 1
41
3 +x 2x
y =
ANSWER
2x5
–15
–
y = –1
y =
ANSWER
ESSENTIAL QUESTION
What are formulas, and how are formulas used?Formulas are equations that
relate two or more quantities, usually represented by
variables. Formulas can be used to solve many real-world
problems, such as problems about investment, temperature,
perimeter, area and volume.
A balloon is released from a height of 5 feet
above the ground. Its
altitude (in feet) after t minutes is
given by the expression
5+82t. What is the altitude of
the balloon after 6 minutes.
LESSON 5: USE PROBLEM SOLVING STRATEGIES AND
MODELS
ESSENTIAL QUESTION
How can problem solving strategies be used to find
verbal and algebraic models?
• Verbal model: a word equation that may be written before an equation is written in mathematical symbols
VOCABULARY
EXAMPLE 1 Use a formula
High-speed Train
The Acela train travels between Boston and Washington, a distance of 457 miles. The trip takes 6.5 hours. What is the average speed?
SOLUTION
You can use the formula for distance traveled as a verbal model.
457 = r 6.5
Distance (miles)
= Rate (miles/hour)
Time (hours)
EXAMPLE 1 Use a formula
An equation for this situation is 457 = 6.5r. Solve for r.
457 = 6.5r
70.3 r
Write equation.
Divide each side by 6.5.
The average speed of the train is about 70.3 miles per hour.
ANSWER
You can use unit analysis to check your answer.
457 miles 6.5 hours70.3 miles
1 hour
CHECK
GUIDED PRACTICE for Example 1
1. AVIATION: A jet flies at an average speed of 540 miles per hour. How long will it take to fly from New York to Tokyo, a distance of 6760 miles?
Jet takes about 12.5 hours to fly from New York to Tokyo.
ANSWER
EXAMPLE 2 Look for a pattern
Paramotoring
A paramotor is a parachute propelled by a fan-like motor. The table shows the height h of a paramotorist t minutes after beginning a descent. Find the height of the paramotorist after 7 minutes.
EXAMPLE 2 Look for a pattern
SOLUTION
The height decreases by 250 feet per minute.
You can use this pattern to write a verbal model for the height.
An equation for the height is h = 2000 – 250t.
EXAMPLE 2 Look for a pattern
So, the height after 7 minutes is h = 2000 – 250(7) = 250 feet.
ANSWER
EXAMPLE 3 Draw a diagram
Banners
You are hanging four championship banners on a wall in your school’s gym. The banners are 8 feet wide. The wall is 62 feet long. There should be an equal amount of space between the ends of the wall and the banners, and between each pair of banners. How far apart should the banners be placed?
SOLUTION
Begin by drawing and labeling a diagram, as shown below.
EXAMPLE 3 Draw a diagram
From the diagram, you can write and solve an equation to find x.
x + 8 + x + 8 + x + 8 + x + 8 + x = 62
5x + 32 = 62
Subtract 32 from each side.5x = 30
x = 6 Divide each side by 5.
Combine like terms.
Write equation.
The banners should be placed 6 feet apart.
ANSWER
EXAMPLE 4 Standardized Test Practice
SOLUTION
STEP 1 Write a verbal model. Then write an equation.
An equation for the situation is 460 = 30g + 25(16 – g).
EXAMPLE 4 Standardized Test Practice
Solve for g to find the number of gallons used on the highway.
STEP 2
460 = 30g + 25(16 – g)
460 = 30g + 400 – 25g
460 = 5g + 400
60 = 5g
12 = g
Write equation.
Distributive property
Combine like terms.
Subtract 400 from each side.
Divide each side by 5.
The car used 12 gallons on the highway.
ANSWER The correct answer is B.
CHECK: 30 12 + 25(16 – 12) = 360 + 100 = 460
GUIDED PRACTICE for Examples 2, 3 and 4
2. PARAMOTORING: The table shows the height h of a paramotorist after t minutes. Find the height of the paramotorist after 8 minutes.
So, the height after 8 minutes is h = 2400 – 210(8) = 720 ft
ANSWER
GUIDED PRACTICE for Examples 2, 3 and 4
3. WHAT IF? In Example 3, how would your answer change if there were only three championship banners?
The space between the banner and walls and between each pair of banners would increase to 9.5 feet.
ANSWER
GUIDED PRACTICE for Examples 2, 3 and 4
4. FUEL EFFICIENCY A truck used 28 gallons of gasoline and traveled a total distance of 428 miles. The truck’s fuel efficiency is 16 miles per gallon on the highway and 12 miles per gallon in the city. How many gallons of gasoline were used in the city?
Five gallons of gas were used.
ANSWER
ESSENTIAL QUESTION
How can problem solving strategies be used to find
verbal and algebraic models?
The problem solving strategies use a formula and look for a pattern that can be used to write verbal models
which can then be used to write algebraic models. The strategy
“draw a diagram” can be used to write an algebraic model directly.
On a blank sheet of paper
complete #2-12 Even on P40 in the blue quiz
section. Turn into the
homework bin when
finished.
LESSON 6: SOLVE LINEAR
INEQUALITIES
ESSENTIAL QUESTION
How are the rules for solving linear inequalities
similar to those for solving linear equations,
and how are they different?
• Linear inequality: an inequality
using <, >, ≥, ≤
• Compound inequality: consists of two simple inequalities joined by “and” or “or”
• Equivalent inequalities: inequalities that have the same solutions as the original inequality
VOCABULARY
• Solve inequalities just the same as equalities using the Order of Operations
• When multiplying or dividing by a negative, flip the inequality sign.
NOTE:
EXAMPLE 1 Graph simple inequalities
a. Graph x < 2.
The solutions are all real numbers less than 2.
An open dot is used in the graph to indicate 2 is not a solution.
EXAMPLE 1 Graph simple inequalities
b. Graph x ≥ –1.
The solutions are all real numbers greater than or equal to –1.
A solid dot is used in the graph to indicate –1 is a solution.
EXAMPLE 2 Graph compound inequalities
a. Graph –1 < x < 2.
The solutions are all real numbers that are greater than –1 and less than 2.
EXAMPLE 2 Graph compound inequalities
b. Graph x ≤ –2 or x > 1.
The solutions are all real numbers that are less than or equal to –2 or greater than 1.
GUIDED PRACTICE for Examples 1 and 2
Graph the inequality.
1. x > –5
The solutions are all real numbers greater than 5.
An open dot is used in the graph to indicate –5 is not a solution.
GUIDED PRACTICE for Examples 1 and 2
Graph the inequality.
2. x ≤ 3
The solutions are all real numbers less than or equal to 3.
A closed dot is used in the graph to indicate 3 is a solution.
GUIDED PRACTICE for Examples 1 and 2
Graph the inequality.
3. –3 ≤ x < 1
The solutions are all real numbers that are greater than or equalt to –3 and less than 1.
GUIDED PRACTICE for Examples 1 and 2
Graph the inequality.
4. x < 1 or x ≥ 2
The solutions are all real numbers that are less than 1 or greater than or equal to 2.
EXAMPLE 3 Solve an inequality with a variable on one side
Fair
You have $50 to spend at a county fair. You spend $20 for admission. You want to play a game that costs $1.50. Describe the possible numbers of times you can play the game.
SOLUTION
STEP 1
Write a verbal model. Then write an inequality.
EXAMPLE 3 Solve an inequality with a variable on one side
An inequality is 20 + 1.5g ≤ 50.
STEP 2 Solve the inequality.
20 + 1.5g ≤ 50
1.5g ≤ 30
g ≤ 20
Write inequality.
Subtract 20 from each side.
Divide each side by 1.5.
ANSWER
You can play the game 20 times or fewer.
EXAMPLE 4 Solve an inequality with a variable on both sides
Solve 5x + 2 > 7x – 4. Then graph the solution.
5x + 2 > 7x – 4
– 2x + 2 > – 4
– 2x > – 6
x < 3
Write original inequality.
Subtract 7x from each side.
Subtract 2 from each side.Divide each side by –2 and reverse the inequality.
ANSWER
The solutions are all real numbers less than 3. The graph is shown below.
GUIDED PRACTICE for Examples 3 and 4
Solve the inequality. Then graph the solution.
5. 4x + 9 < 25
6. 1 – 3x ≥ –14
7. 5x – 7 ≤ 6x
8. 3 – x > x – 9
x < 4
ANSWER
x ≤ 5
ANSWER
x < 6
ANSWER
x > – 7
ANSWER
EXAMPLE 5 Solve an “and” compound inequality
Solve – 4 < 6x – 10 ≤ 14. Then graph the solution.
– 4 < 6x – 10 ≤ 14
– 4 + 10 < 6x – 10 + 10 ≤ 14 + 10
6 < 6x ≤ 24
1 < x ≤ 4
Write original inequality.
Add 10 to each expression.
Simplify.
Divide each expression by 6.
ANSWER
The solutions are all real numbers greater than 1 and less than or equal to 4. The graph is shown below.
EXAMPLE 6 Solve an “or” compound inequality
Solve 3x + 5 ≤ 11 or 5x – 7 ≥ 23 . Then graph the solution.
SOLUTION
A solution of this compound inequality is a solution of either of its parts.
First Inequality Second Inequality
3x + 5 ≤ 11
3x ≤ 6
x ≤ 2
Write first inequality.
Subtract 5 from each side.
Divide each side by 3.
5x – 7 ≥ 23
5x ≥ 30
x ≥ 6
Write second inequality.
Add 7 to each side.
Divide each side by 5.
EXAMPLE 6 Solve an “or” compound inequality
ANSWER
The graph is shown below. The solutions are all real numbers less than or equal to 2 or greater than or equal to 6.
EXAMPLE 7 Write and use a compound inequality
Biology
A monitor lizard has a temperature that ranges from 18°C to 34°C. Write the range of temperatures as a compound inequality. Then write an inequality giving the temperature range in degrees Fahrenheit.
EXAMPLE 7 Write and use a compound inequality
SOLUTION
The range of temperatures C can be represented by the inequality 18 ≤ C ≤ 34. Let F represent the temperature in degrees Fahrenheit.
18 ≤ C ≤ 34 Write inequality.
18 ≤ ≤ 345
9(F – 32)
32.4 ≤ F – 32 ≤ 61.2
64.4 ≤ F ≤ 93.2
Substitute for C.95 (F – 32)
Multiply each expression by ,
the reciprocal of .
955
9Add 32 to each expression.
EXAMPLE 7 Write and use a compound inequality
ANSWER
The temperature of the monitor lizard ranges from 64.4°F to 93.2°F.
GUIDED PRACTICE for Examples 5,6, and 7
Solve the inequality. Then graph the solution.
9. –1 < 2x + 7 < 19
ANSWER
The solutions are all real numbers greater than – 4 and less than 6.
–4 < x < 6
GUIDED PRACTICE for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
10. –8 ≤ –x – 5 ≤ 6
The solutions are all real numbers greater than and equal to – 11 and less than and equal to 3.
ANSWER
–11 ≤ x ≤ 3
GUIDED PRACTICE for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
11. x + 4 ≤ 9 or x – 3 ≥ 7
ANSWER
The graph is shown below. The solutions are all real numbers.
less than or equal to 5 or greater than or equal to 10.
x ≤ 5 or x ≥ 10
GUIDED PRACTICE for Examples 5,6 and 7
Solve the inequality. Then graph the solution.
12. 3x – 1< –1 or 2x + 5 ≥ 11
x < 0 or x ≥ 3
less than 0 or greater than or equal to 3.
ANSWER
The graph is shown below. The solutions are all real numbers.
GUIDED PRACTICE for Examples 5,6 and 7
13. WHAT IF? In Example 7, write a compound inequality for a lizard whose temperature ranges from 15°C to 30°C. Then write an inequality giving the temperature range in degrees Fahrenheit.
15 ≤ C ≤ 30 or 59 ≤ F ≤ 86
ANSWER
ESSENTIAL QUESTION
How are the rules for solving linear inequalities similar to
those for solving linear equations, and how are they
different?The addition and subtraction
properties are the same, but if you multiply or divide both sides
of an inequality by a negative number, the inequality symbol
must be reversed.
What is unique about
absolute value
numbers that is not true of
integers?
LESSON 7: SOLVE ABSOLUTE VALUE EQUATIONS AND
INEQUALITIES
ESSENTIAL QUESTION
How are absolute value equations and inequalities like linear equations and
inequalities?
• Absolute value: the distance an umber is from 0 on a number line; always positive
• Extraneous solution: an apparent solution that must be rejected because it does not satisfy the original equation
VOCABULARY
EXAMPLE 1 Solve a simple absolute value equation
Solve |x – 5| = 7. Graph the solution.
SOLUTION
| x – 5 | = 7
x – 5 = – 7 or x – 5 = 7
x = 5 – 7 or x = 5 + 7
x = –2 or x = 12
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
EXAMPLE 1
The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.
ANSWER
Solve a simple absolute value equation
EXAMPLE 2 Solve an absolute value equation
| 5x – 10 | = 45
5x – 10 = 45 or 5x – 10 = –45
5x = 55 or 5x = –35
x = 11 or x = –7
Write original equation.
Expression can equal 45 or –45 .
Add 10 to each side.
Divide each side by 5.
Solve |5x – 10 | = 45.
SOLUTION
EXAMPLE 2 Solve an absolute value equation
The solutions are 11 and –7. Check these in the original equation.
ANSWER
Check:
| 5x – 10 | = 45
| 5(11) – 10 | = 45?
|45| = 45?
45 = 45
| 5x – 10 | = 45
| 5(–7) – 10 | = 45?
45 = 45
| – 45| = 45?
EXAMPLE 3
| 2x + 12 | = 4x
2x + 12 = 4x or 2x + 12 = – 4x
12 = 2x or 12 = –6x
6 = x or –2 = x
Write original equation.
Expression can equal 4x or – 4 x
Add –2x to each side.
Solve |2x + 12 | = 4x. Check for extraneous solutions.
SOLUTION
Solve for x.
Check for extraneous solutions
EXAMPLE 3
| 2x + 12 | = 4x
| 2(–2) +12 | = 4(–2)?
|8| = – 8?
8 = –8
Check the apparent solutions to see if either is extraneous.
Check for extraneous solutions
| 2x + 12 | = 4x
| 2(6) +12 | = 4(6)?
|24| = 24?
24 = 24
The solution is 6. Reject –2 because it is an extraneous solution.
ANSWER
CHECK
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
1. | x | = 5
for Examples 1, 2 and 3
The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1
2
3
4
5
6
7
– 5
– 6
– 7
5 5
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
2. |x – 3| = 10
for Examples 1, 2 and 3
The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1
2
3
4
5
6
7
– 5
– 6
– 7
8
9
10
11
12
13
10 10
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
3. |x + 2| = 7
for Examples 1, 2 and 3
The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.
ANSWER
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
4. |3x – 2| = 13
for Examples 1, 2 and 3
ANSWER
The solutions are 5 and .
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
5. |2x + 5| = 3x
for Examples 1, 2 and 3
The solution of is 5. Reject 1 because it is an extraneous solution.
ANSWER
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
6. |4x – 1| = 2x + 9
for Examples 1, 2 and 3
ANSWER
The solutions are – and 5. 3
11
EXAMPLE 4 Solve an inequality of the form |ax + b| > c
Solve |4x + 5| > 13. Then graph the solution.
SOLUTION
First Inequality Second Inequality
4x + 5 < –13 4x + 5 > 13
4x < –18 4x > 8
x < – 92
x > 2
Write inequalities.
Subtract 5 from each side.
Divide each side by 4.
The absolute value inequality is equivalent to
4x +5 < –13 or 4x + 5 > 13.
EXAMPLE 4
ANSWER
Solve an inequality of the form |ax + b| > c
The solutions are all real numbers less than or greater than 2. The graph is shown below.– 9
2
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
7. |x + 4| ≥ 6
x < –10 or x > 2 The graph is shown below.
ANSWER
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
8. |2x –7|>1
ANSWER
x < 3 or x > 4 The graph is shown below.
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
9. |3x + 5| ≥ 10
ANSWER
x < –5 or x > 123
The graph is shown below.
EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c
A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.
Baseball
SOLUTION
Write a verbal model. Then write an inequality.STEP 1
EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c
STEP 2 Solve the inequality.
Write inequality.
Write equivalent compound inequality.
Add 5.125 to each expression.
|w – 5.125| ≤ 0.125
– 0.125 ≤ w – 5.125 ≤ 0.125
5 ≤ w ≤ 5.25
So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.
ANSWER
EXAMPLE 6
The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.
Gymnastics
SOLUTION
STEP 1 Calculate the mean of the extreme mat thicknesses.
Write a range as an absolute value inequality
EXAMPLE 6
Mean of extremes = = 7.875 7.5 + 8.25 2
Find the tolerance by subtracting the mean from the upper extreme.
STEP 2
Tolerance = 8.25 – 7.875
Write a range as an absolute value inequality
= 0.375
EXAMPLE 6
STEP 3 Write a verbal model. Then write an inequality.
A mat is acceptable if its thickness t satisfies |t – 7.875| ≤ 0.375.
ANSWER
Write a range as an absolute value inequality
GUIDED PRACTICE for Examples 5 and 6
Solve the inequality. Then graph the solution.
10. |x + 2| < 6
The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.
ANSWER
–8 < x < 4
GUIDED PRACTICE for Examples 5 and 6
Solve the inequality. Then graph the solution.
11. |2x + 1| ≤ 9
The solutions are all real numbers less than –5 or greater than 4. The graph is shown below.
ANSWER
–5 ≤ x ≤ 4
GUIDED PRACTICE for Examples 5 and 6
12. |7 – x| ≤ 4
Solve the inequality. Then graph the solution.
3 ≤ x ≤ 11
ANSWER
The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.
GUIDED PRACTICE for Examples 5 and 6
13. Gymnastics: For Example 6, write an absolute value inequality describing the unacceptable mat thicknesses.
A mat is unacceptable if its thickness t satisfies |t – 7.875| > 0.375.
ANSWER
ESSENTIAL QUESTION
How are absolute value equations and inequalities like linear equations and
inequalities?An absolute value equation
can be rewritten as two linear equations, and an
absolute value inequality can be rewritten as two linear
inequalities.