Transcript 45 watt

Solving Two-Step Equations
LESSON 6-1
Course 3
Problem of the Day
There are 15 girls in a class of 27 students. Using lowest terms, what fraction
of the students are girls? What fraction are boys?
5 4
,
9 9
Lesson
Main
Lesson
6-1
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-6.)
1. Vocabulary Review What does it mean to isolate the variable?
Solve each equation.
2. x + 4 = – 3
3. c – 5 = 1
4. 5 + a = 35
Check Skills You’ll Need
Lesson
Main
Lesson
6-1
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Check Skills You’ll Need
Solutions
1. to get the variable alone on one side of the equation
2.
x+4=–3
x+4–4 =–3–4
x=–7
3.
c–5= 1
c–5+5= 1+5
c=6
4. a = 30
Lesson
Main
Lesson
6-1
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Additional Examples
Solve 4p + 27 = 61.48.
4p + 27 = 61.48
4p + 27 – 27 = 61.48 – 27
4p = 34.48
34.48
4p
=
4
4
Subtract 27 from each side.
Simplify.
Divide each side by 4.
Simplify.
p = 8.62
Check 4p + 27 = 61.48
61.48
Substitute 8.62 for p.
61.48 = 61.48
The solution checks.
4(8.62) + 27
Quick Check
Lesson
Main
Lesson
6-1
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Additional Examples
At a recent breakfast, four friends paid for their
drinks and shared the cost of a bag of doughnuts. Joe’s
drink was $1.75. He paid $3.20 total for breakfast. What
equation can be used to find the cost of the doughnuts?
How much did the bag of doughnuts cost?
Words
cost of
Joe’s drink
(cost of bag of
doughnuts 4)
plus
is
$3.20
=
3.20
Let d = the cost of a bag of doughnuts.
Equation
1.75
Lesson
Main
d
4
+
Lesson
6-1
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Additional Examples
(continued)
The equation is 1.75 + d =3.2. You can solve the equation to find the
4
cost.
1.75 + d = 3.2.
4
1.75 – 1.75 + d
4
d
4
(4) d
4
= 3.2 – 1.75
Subtract 1.75 from each side.
= 1.45
Simplify.
= (4)1.45
Multiply each side by 4.
d = 5.8
Simplify.
The cost of a bag of doughnuts is $5.80.
Lesson
Main
Lesson
6-1
Quick Check
Feature
Solving Two-Step Equations
LESSON 6-1
Course 3
Lesson Quiz
Solve each equation.
1.
n
– 14 = 1
3
45
2. 16 – 2x = 9
3.5
Write and solve an equation.
3. Suppose you bought a $2.75 sandwich and two drinks of equal
price. You spend $4.25 in all. How much does one drink cost?
2d + 2.75 = 4.25; $.75
4. Admission to a museum costs $3.75 per person. A group of friends
attend. They spend a total of $30 on lunch. At the end of the day, they
had spent a total of $56.25. How many friends attended the museum?
3.75x + 30 = 56.25; 7 friends
Lesson
Main
Lesson
6-1
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Problem of the Day
Suppose you had the following set of numbers:
2, 3, 5, 7, 8, 11
Using only two numbers at a time, write as many fractions greater than 1 as
possible from this set.
3 , 5 , 7 , 8 , 11 , 5 , 7 , 8 , 11 , 7 , 8 , 11 , 8 , 11 , 11
2 2 2 2 2 3 3 3 3 5 5 5 7 7 8
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-5.)
1. Vocabulary Review Is the expression (5 + 3a) – 15 simplified? Explain.
Simplify each expression.
2. – 8(r + 3)
4. 35(2 – t)
3. – 7(s – 5)
Check Skills You’ll Need
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Check Skills You’ll Need
Solutions
1. No; 15 can be subtracted from 5. The simplest form is 3a – 10.
2. – 8(r) + (– 8)(3) = – 8r – 24
3. –7(s) – (– 7)(5) = – 7s + 35
4. 35(2) – (35)(t) = 70 – 35t
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Additional Examples
Combine like terms in the expression
8p + 13p + p.
8p + 13p + p = 8p + 13p + 1p
Rewrite p as 1p.
= (8 + 13 + 1)p
Distributive Property
= 22p
Combine like terms by adding.
Quick Check
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Additional Examples
Carlos buys 6 tubes of paint and 3 pieces of
fabric to make an art project. Shauna buys 8 tubes of paint
and 1 piece of fabric. Define and use variables to represent
the total cost.
Words
Carlos
6 tubes of paint
plus
3 pieces of fabric
Let t = the cost of a tube of paint.
Let f = the cost of a piece of fabric.
Expression
Words Shauna
Expression
Lesson
Main
6t
3f
+
8 tubes of paint
plus
+
8t
Lesson
6-2
1 piece of fabric
f
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Additional Examples
(continued)
Combined Expression (6t + 3f ) + (8t + f )
Commutative Property
of Addition
(6t + 3f) + (8t + f) = 6t + 3f + 8t + f
= (6 + 8)t + (3 + 1)f
Distributive Property
= 14t + 4f
Simplify.
Quick Check
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Additional Examples
Simplify 7t – 2(t – 3).
7t – 2(t – 3) = 7t + (–2)(t – 3)
Add the opposite of 2(t – 3).
= 7t + [–2t – (–6)]
Distributive Property
= 7t + (–2t) + 6
Simplify.
= [7 + (–2)]t + 6
Distributive Property
= 5t + 6
Simplify.
Quick Check
Lesson
Main
Lesson
6-2
Feature
Simplifying Algebraic Expressions
LESSON 6-2
Course 3
Lesson Quiz
Simplify each expression.
1. – 13c + c
–12c
2. 4y – 7 + 8y
12y – 7
3. 1 – 6(b – 9)
– 6b + 55
4. Karen buys 4 boxes of cereal and 3 bags of almonds at the grocery
store. Her brother, David, buys 2 boxes of cereal. Define and use
variables to represent the total cost.
Let c = the cost of a box of cereal and let a = the cost of a bag of
almonds. Then 6c + 3a represents the total cost.
Lesson
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Lesson
6-2
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Problem of the Day
Find the lowest common denominator (LCD) for each set of numbers.
a. 1 , 3 , 1
2
4 3
b. 2 , 5 , 1
3
12
6
9
18
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Check Skills You’ll Need
(For help, go to Lesson 6-2.)
1. Vocabulary Review Identify the like terms in 3x + 2x = 8 – x.
Simplify.
2. 5 – 3m + 7 – 23m
3. 4(7 – 3r)
4. (q + 1)5 + 3q
Check Skills You’ll Need
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Check Skills You’ll Need
Solutions
1. 3x, 2x, – x
2. 12 – 26m
3. 4(7) + (4)(– 3r) = 28 – 12r
4. 5(q) + 5(1) + 3q = (5 + 3)q + 5 = 8q + 5
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Additional Examples
Solve 2c + 2 + 3c = 12.
2c + 2 + 3c = 12
2c + 3c + 2 = 12
5c + 2 = 12
5c + 2 – 2 = 12 – 2
5c = 10
5c
10
=
5
5
c=2
Lesson
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Commutative Property
Combine like terms.
Subtract 2 from each side.
Simplify.
Divide each side by 5.
Simplify.
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Additional Examples
(continued)
Check 2c + 2 + 3c = 12
2(2) + 2 + 3(2)
12
12 = 12
Substitute 2 for c.
The solution checks.
Quick Check
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Additional Examples
Eight cheerleaders set a goal of selling 424 boxes of
cards to raise money. After two weeks, each cheerleader has sold
28 boxes. How many more boxes must each cheerleader sell?
Words 8 cheerleaders • (28 boxes +
additional
boxes
= 424 boxes
Let x = the number of additional boxes.
Equation
8
•
(28
+
x)
=
424
8(28 + x) = 424
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Additional Examples
Quick Check
(continued)
224 + 8x = 424
Distributive Property
224 – 224 + 8x = 424 – 224
Subtract 224 from each side.
8x = 200
Simplify.
8x
200
=
8
8
Divide each side by 8.
x = 25
Simplify.
Each cheerleader must sell 25 more boxes.
Check for Reasonableness Round 8 to 10 and 28 to 20. The cheerleaders
sold about 10 • 20, or 200 boxes. This means each cheerleader must sell
about 22 more boxes. 25 is close to 22. The answer is reasonable.
Lesson
Main
Lesson
6-3
Feature
Solving Multi-Step Equations
LESSON 6-3
Course 3
Lesson Quiz
Solve the following equations.
1. 2m + 4 – 8m = 28
m = –4
2. 2(f – 1) + f = 37
f = 13
3. 4.5(4x – 12) = 144
x = 11
4. Jasmine earns a certain amount per hour for the first 40 hours worked in a
week. Each hour she works over 40 in one week, she earns an additional
$4.50 per hour. If Jasmine works 46 hours one week, and earned $383.50
that week, how much does she earn per hour for the first 40 hours?
$7.75 per hour
Lesson
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Lesson
6-3
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Problem of the Day
Audra drinks 24.75 oz of milk in 1 day. How many ounces of milk does she
drink in a week?
173.25 oz
Lesson
Main
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Check Skills You’ll Need
(For help, go to Lesson 6-2.)
1. Vocabulary Review Operations that undo each other are called
Simplify.
2. 9(t + 7) – 16
3. 12 – 6(2r – 8)
4. 2x – (5x + 7)
Check Skills You’ll Need
Lesson
Main
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Check Skills You’ll Need
Solutions
1. inverse operations
2. 9t + 47
3. 60 – 12r
4. – 3x – 7
Lesson
Main
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Additional Examples
Solve 9 + 2p = –3 – 4p.
9 + 2p = –3 – 4p
9 + 2p + 4p = –3 – 4p + 4p
9 + 6p = –3
9 – 9 + 6p = –3 – 9
Add 4p to each side.
Combine like terms.
Subtract 9 from each side.
6p = –12
Simplify.
6p
–12
=
6
6
Divide each side by 6.
p = –2
Lesson
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Simplify.
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Additional Examples
(continued)
Check 9 + 2p = –3 – 4p
9 + 2(–2)
–3 – 4(–2)
5=5
Substitute –2 for p.
The solution checks.
Quick Check
Lesson
Main
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Additional Examples
Each week you set aside $18 for a stereo and put
the remainder in a savings account. After 7 weeks, the amount
you place in the savings account is 4.2 times your total weekly
pay. How much do you make each week?
Words
7
(weekly amount – 18)
= 4.2
Weekly amount
Let x = the amount earned weekly.
7
Lesson
Main
Lesson
6-4
(x - 18) = 4.2
x
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Additional Examples
(continued)
7(x – 18) = 4.2x
7x – 126 = 4.2x
7x – 7x – 126 = 4.2x – 7x
Distributive Property
Subtract 7x from each side.
– 126 = – 2.8x
Combine like terms.
– 126 – 2.8x
=
– 2.8
– 2.8
45 = x
Divide each side by – 2.8.
Simplify.
Quick Check
Lesson
Main
Lesson
6-4
Feature
Solving Equations With Variables on Both Sides
LESSON 6-4
Course 3
Lesson Quiz
Solve each equation.
1. 4(3u – 1) = 20.
u=2
2. 5t – 4 = t – 8.
t = –1
3. 3(k – 8) = – k.
k=6
Lesson
Main
Lesson
6-4
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Problem of the Day
If 1.5 truckloads of dirt cost $39.00, what is the cost of 1 truckload?
$26.00
Lesson
Main
Lesson
6-5
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-6.)
1. Vocabulary Review A(n) ? is a mathematical sentence
with an equal sign.
Solve each equation.
2. x + 15 = – 3
3. y + 22 = 9
4. a – 28 = – 4
Check Skills You’ll Need
Lesson
Main
Lesson
6-5
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Check Skills You’ll Need
Solutions
1.
equation
2.
x + 15 = – 3
x + 15 – 15 = – 3 – 15
x = – 18
4.
a – 28 = – 4
a – 28 + 28 = – 4 + 28
a = 24
Lesson
Main
3.
Lesson
6-5
y + 22 = 9
y + 22 – 22 = 9 –22
y = – 13
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Additional Examples
Solve p – 3 < –5.
p – 3 < –5
p – 3 + 3 < –5 + 3
p < –2
Notice that 3 is subtracted from p.
Add 3 to each side.
Simplify.
Check
Step 1 See if –2 is a solution to the related equation.
p – 3 = –5
–2 – 3
–5
–5 = –5
Lesson
Main
Write the related equation.
Substitute –2 for p.
Simplify.
Lesson
6-5
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Additional Examples
(continued)
Step 2 Check the inequality symbol. Choose any number less than –2 and
substitute it into the original inequality. In this case, try –3.
p – 3 < –5
–3 – 3 < –5
Substitute –3 for p.
–6 < –5
Steps 1 and 2 both check, so p < –2 is the solution of p – 3 < –5.
Quick Check
Lesson
Main
Lesson
6-5
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Additional Examples
After the hairdresser cut 3 in. from Rapunzel’s hair, her hair
was at least 15 in. long. How long was her hair before she had it cut?
Words length before cut – length cut is at least 15 in.
Let l = the length before cut.
Inequality
l–3 >
– 15
l–3+3 >
– 15 + 3
l >
– 18
l
–
3
>
–
15
Notice that 3 is subtracted from l.
Add 3 to each side.
Simplify.
Rapunzel’s hair was at least 18 in. long before she had it cut.
Lesson
Main
Lesson
6-5
Quick Check
Feature
Solving Inequalities by Adding or Subtracting
LESSON 6-5
Course 3
Lesson Quiz
1. Solve r – 9 > 8.
r > 17
2. Solve and graph the inequality: x + 6 > 4.
x > –2
3. Solve and graph the inequality: x – 14 < – 13.
x<1
4. A lamp can use lightbulbs of up to 75 watts. The lamp is using a 45-watt bulb.
At most, how many watts are available for brighter light?
30 watts
Lesson
Main
Lesson
6-5
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Problem of the Day
What is the greatest common factor (GCF) of 60 and 72?
12
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-7.)
1. Vocabulary Review Is the definition “Negative numbers are numbers less
than or equal to zero” correct? Explain.
Solve each equation.
2. 4x = – 16
3. – 8p = 808
4. – 2u = – 12.4
t
5. 1 = –8
Check Skills You’ll Need
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Check Skills You’ll Need
Solutions
1. No; zero is not a negative number.
2. 4x = –16
4x = –16
4
4
x= –4
5.
1=
3. –8p = 808
–8p
= 808
8
–8
p = –101
4. –2u = –12.4
–2u
–12.4
=
–2
–2
u = 6.2
t
–8
(–8) · 1 = (–8) · (
t
)
–8
–8 = t
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Additional Examples
A small business sells each CD of its game software for $12.
How many CDs must they sell to meet the goal of at least $84,000?
Words number of CDs • $12 is at least $84,000
Let c = the number of CDs.
Inequality
c
• 12
>
–
84,000
12c >
– 84,000
12c > 84,000
12 –
12
c >
– 7,000
Lesson
Main
Divide each side by 12.
Simplify.
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Additional Examples
(continued)
The business must sell at least 7,000 CDs.
Check for Reasonableness The answer makes sense because 7,000 • 12
is 84,000, and any number over 7,000 multiplied by 12 is a number greater
than 84,000.
Quick Check
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Additional Examples
Solve
y
> 3. Graph the solution.
–4
y
>3
–4
–4•
y
–4
<–4•3
y < –12
Multiply each side by – 4. Reverse the direction of
the inequality.
Simplify.
Quick Check
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Additional Examples
Solve –5b <
– 15. Graph the solution.
–5b <
– 15
–5b > 15
–5 – –5
b >
– –3
Divide each side by –5. Reverse the
direction of the inequality.
Simplify.
Quick Check
Lesson
Main
Lesson
6-6
Feature
Solving Inequalities by Multiplying or Dividing
LESSON 6-6
Course 3
Lesson Quiz
Solve each inequality.
2. – 7r ≥ 14
1. 9c ≤ –36
c  –4
3.
r –2
a
<–1
–8
a>8
4. Leroy’s class is having a cookie sale to raise money for a class trip. They will
sell homemade cookies for $1.50 each. If they earn at least $200, a local business
will match the amount they earn. At least how many cookies do they need to sell to
get the local company to match their cookie-sale earnings?
134 cookies
Lesson
Main
Lesson
6-6
Feature