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Order of
Operations
with
rational
numbers
objective
 use the order of
operation to simplify
numerical expression
containing rational
numbers
Example 1.
Simplify: (-0.4)3(-5/2)2
(-0.4)3(-5/2)2
(-0.4)(-0.4)(-0.4) (-5/2)(-.5/2)
-0.064 25/4
Simplify each power
before multiplying
each factor.
Multiply like
terms.
Multiply.
-0.064 25/4
-1.6/4
Multiply the factors.
Simplify.
-0.4
To simplify expressions that contain more than
one grouping symbol, begin computing with the
innermost set
Example 2.
Simplify: -6(2/3- 5/9) ÷ [(2.4 5)(-1)5]
-6(2/3-5/9) ÷ [(2.4
5)(-1)5]
-6(1/9) ÷ [(12)(-1)5]
-6(1/9) ÷ [(12)(-1)]
Begin computing
within parentheses.
Simplify the power.
Multiply.
-6(1/9) ÷ [(12)(-1)]
Multiply.
-6/9 ÷ (-12)
Simplify.
-2/3 ÷ (-12)
Divide; multiply
by the reciprocal.
-2/3 -1/12
2/36
= 1/18
Simplify.
The division bar is a grouping symbol. To work on
an expression with a division bar, first simplify the
numerator, then the denominator, and finally divide.
Example 3
52 – 7 2/10
Simplify:
(2 – 15) + 23
52 – 7 2/10
(2 – 15) + 23
Simplify the power.
25 – 7 2/10
(2 – 15) + 23
Subtract and rewrite the
answer in simplest form.
25 – 7 2/10
(2 – 15) + 23
Subtract and rewrite the
answer in simplest form.
17 4/5
(2 – 15) + 23
Compute within parentheses.
17 4/5
-13 + 23
17 4/5
-13 + 8
17 4/5
-5
Simplify the power.
Add.
Divide.
17 4/5
-5
17 4/5 ÷ (-5)
Divide.
Rewrite in horizontal
form.
17 4/5 ÷ (-5)
Rename as fractions.
89/5 ÷ -5/1
Multiply by the
reciprocal to divide
89/5 x -1/5
Rename as a mixed
number
-89/25
-3 14/25
The division bar is a grouping symbol. To work on
an expression with a division bar, first simplify the
numerator, then the denominator, and finally divide.
Example 4
-12.5 + 0.5
Simplify:
3/4 0.5
-12.5 + 0.5
3/4 0.5
-12
3/4 0.5
Add
Rename 0.5 as 1/2.
-12
3/4 0.5
-12
3/4 1/2
-12
3/8
-12 ÷ 3/8
-12/1 x 8/3
Rename 0.5 as 1/2.
Multiply.
Simplify. Write in
horizontal form.
Write as multiplication.
Simplify.
-4/1 x 8/1 = -32/1 = -32
Homework
PB, p 147-148
Addition/Subtraction Equations With Fractions
Example 3. Solve and check.
x – 2 5/8 = 1 1/4
x – 2 5/8 + 2 5/8 = 1 1/4 + 2 5/8
x = 1 2/8 + 2 5/8
x = 3 7/8
Substitute 3 7/8 for x to check.
x – 2 5/8 = 1 1/4
3 7/8 – 2 5/8 = 1 1/4
Simplify.
1 2/8 = 1 1/4
1 1/4 = 1 1/4; true
3 7/8 is a solution.
Homework
PB, p 149-150
Addition/Subtraction Equations With Fractions
Example 2. Solve and check.
-5/8 – 1/8 + n = 1
-6/8 + n = 1
+6/8
+6/8
n = 1 6/8
n = 1 3/4
Combine like terms.
Simplify. Add 6/8
to both sides.
Simplify the fraction
Check the solution.
Replace n with 1 3/4
-5/8 – 1/8 + 1 3/4 = 1
-6/8 + 1 3/4 = 1
Addition/Subtraction Equations With Fractions
Example 2. Solve and check.
n = 1 3/4
Check the solution.
-5/8 – 1/8 + n = 1 Replace n with 1 3/4.
Combine
-5/8 – 1/8 + 1 3/4 = 1
Simplify
-6/8 + 1 3/4 = 1
Add.
-3/4 + 1 3/4 = 1
1 = 1 True, so 1 3/4 is a solution.
Multiplication and
division equations
with fractions
Text, pp 136-137
objective:
 apply the Multiplication
Property of Equality
Example 1. Solve and check.
1/4 w + 2/4 w = 15 Combine like terms.
Multiply both
4/3 3/4 w = 15 4/3
sides by 4/3.
w = 60/3
Divide.
w = 20
1/4 (20) + 2/4 (20) =
5 + 10 = 15
Check. Substitute
20 for w.
Simplify.
15
True. So 20 is a solution
Homework
PB, p 151-152
Two-Step equations
with fractions
Text, pp 138-139
objective:
 apply the properties of
equality to simplify twostep equations with fractions
Example 1. Solve and check.
Add 16 ½ to both
1/2 p –16 1/2 = 15
+16 1/2 +16 1/2 sides.
2/1 1/2 p = 31 1/2 2/1 Multiply both
p = 31 1/2 2/1
sides by 2/1.
Rename 31 1/2 as
improper fraction
p = 63/2 2/1 = 63/1 = 63
1/2 p –16 1/2 = 15
1/2 63 –16 1/2 = 15
Check.
Substitute 63 for p.
Example 1. Solve and check.
1/2 p –16 1/2 = 15 Substitute 63 for p.
1/2 63 –16 1/2 = 15 Multiply.
31 1/2 –16 1/2 = 15 Subtract.
15 = 15
True, so 63 is
a true solution
Example 2. Solve and check.
d
(-17)
59 =
2 1/4
d
59 =
+17
2
1/4
-17
-17
d
42 =
2 1/4
d 9/4
9/4 42 = 9/4
9/4 42 = d
Simplify the
grouping symbols.
Subtract 17 from
both sides.
Rename 2 1/4 as a
fraction
Multiply both
sides by 4/9
Multiply
Example 2. Solve and check.
2
9/4 42 = d
Simplify.
9/2 21 = d
Multiply.
21
189/2 = d
Rename as mixed number.
Check.
94 1/2 = d
d
Use 94 1/2 in
(- 17)
59 =
2 1/4
place of d.
Example 2. Solve and check.
d
Use
94
1/2
in
59
(- 17)
2 1/4
place of d.
94 1/2
Simplify the
59 =
(- 17) parentheses.
2 1/4
94 1/2
59 =
+ 17 Write the division
2 1/4
in horizontal form.
59 = 94 1/2 ÷ 2 1/4 + 17 Rename as
59 = 189/2 ÷ 9/4 + 17
fractions.
Write as
multiplication.
Example 2. Solve and check.
21
59 = 189/2 4/9 + 17
1
1
59 = 21/2
Simplify.
4/1 + 17
Simplify.
2
Multiply
59 = 21 2 + 17
Add.
59 = 42 + 17
True. So 94 1/2 is a solution.
59 = 59
Class work
PB, p 153
Homework
PB, p 153-154
Customary units of
measure
Text, pp 138-139
objective:
 rename customary units
measure to a larger or
smaller units
 Customary units of
length
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 ft or 36 in
1 mile (mi) = 5280 ft or 1760 yd
 Customary units of
capacity
1 cup (c) = 8 fluid ounces (fl oz)
1 pint (pt) = 2 c
1 quart (qt) = 2 pt
1 gallon (gal) = 4 qt
 Customary units of
weight
1 pound (lb) = 16 ounces (oz)
1 ton (T) = 2000 lb
 Customary units of
Measure
To rename larger units as
smaller units, multiply by the
conversion unit
To rename smaller units as
larger units, divide by the
conversion unit
Example. How many yards are there in 2
½ miles?
Think.  2 1/2 mi = _________ yd
1 mi = 1760 yd
 mi larger than yard
larger to smaller, multiply
2 1/2 mi 1760 yd
5/2 mi 1760 yd
1
880
5 mi 880 yd
4400 yd
Rename as fraction.
Simplify.
Multiply.
2 1/2 mi.
Class work
PB, p 155
Homework
PB, p 155-156
Problem solving strategy:
Make a drawing
Text, pp 138-139
objective:
 solve word problems using the
strategy “Make A Drawing”
Sample Problem 1.
The clock tower in Liberty Square,
known for its accuracy, chimes its bell
every hour on the hour at equal
intervals. If the clock strikes 6 chimes in
6 seconds, how long would it take for the
clock to strike 12 chimes at 12 o’clock?
(To complete the problem, assume that
the chime itself takes no time)
Hint: 12 seconds is not the answer.
Read
Read to understand what is being
asked. (List the facts and restate
the question.)
 Chime occurs in equal
intervals.
 6 chimes strike in 6 seconds
at 6 o’clock.
 The answer is not twelve
seconds.
Question:  How long would it take
for the clock to strike 12
chimes at 12 o’clock.
Facts:
Plan
Select a strategy.
Problem-Solving Strategies
 Guess and test.  Make a drawing.
 Organize data.  Reason logically
 Find a pattern.  Work backward
 Solve a simpler problem.
 Adopt a different point of view.
 Account for all possibilities.
 Consider extreme cases.
Using the strategy “Make a Drawing”
will help you understand the situation.
Solve
Apply the strategy.
First make a drawing that help you
understand the situation. Use dots to
show the chimes that occur at 6 o’clock.
6 sec
1
2
1
3
2
4
3
5
4
6
5
The 6 chimes occur in 6 seconds. There
are 5 intervals in those 6 chimes,
therefore each interval must be 6/5
seconds. Think: 6/5 5 = 6.
Solve
Apply the strategy.
Now make a drawing to show the
situation at 12 o’clock. Use dots also to
show the chimes.
1
2
1
3
2
4
3
5
4
6
5
7
6
8
7
9
8
10
9
11
10
12
11
There are 11 intervals between the 12 chimes
at 12 o’clock. If an interval is 6/5 of a second,
then 6/5 11 will give us what it will take for
the 12 chimes the clock will make at twelve.
Solve
Apply the strategy.
There are 11 intervals between the 12 chimes
at 12 o’clock. If an interval is 6/5 of a second,
then 6/5 11 will give us the it will take for
the 12 chimes the clock will make at twelve.
6/5 11 = 66/5 = 13 1/5 seconds
The clock takes13 1/5 seconds to strike 12
chimes
Check to make sure your answer
Check makes sense
There are twice as many chimes, so it ought
to take twice as long. It appears to be so.
Check
Check to make sure your answer
makes sense
• The 6 chimes occur in 6 seconds.
•The 12 chimes occur in 13 1/5 seconds.
• There are 5 intervals between the 6 chimes.
• There are 11 intervals between the 12
chimes.
There are twice as many chimes, so it ought
to take twice as long. It appears to be so.
There are more than twice as many intervals,
so it ought to take more than twice as long. It
appears to be so.
Sample Problem 2.
There are 240 seven graders at
Kingston Middle School. Of these
students, 1/6 walk to school. Of
those who do not walk, 3/4 take
the bus to school. Of those who do
not walk or take the bus half ride
their bikes. How many seventh
graders ride their bikes to school?
Read
Read to understand what is being
asked. (List the facts and restate
the question.)
Facts:  There are 240 seventh graders in
all
 1/6 walk to school.
 3/4 of those who do not walk
take the bus
 1/2 of those who do not walk or
take the bus ride their bike.
Question:  How many seventh graders
ride their bike to school?.
Plan
Select a strategy.
This problem has a lot of information. To
make this information easier to understand,
you can use the strategy “Make a Drawing”.
Solve
Apply the strategy.
 Draw a rectangle to represent the entire
seventh grade.
 Divide the rectangle to show those who walk
and those who do not.
240
walks
1/6
40
Think. 1/6 of 240 is 40.
do not walk
200
 Divide the section representing those
who do not walk into fourths.
walks
do not walk
50
50
50
50
40
200
Think. 1/4 of 200 is 50
Divide the remaining fourth into two.
50
50
40
walks
50
25
25
50
do not walk or take the bus
So 25 students ride their bikes to school.
Think. 1/2 of 50 is 25.
Check
Check to make sure your answer
makes sense.
Look back at the final drawing. Make sure the
numbers that represent each section satisfy the
condition in the problem.
 The total is 40 + 50 + 50 + 50 + 25 + 25 =
240.
 40 students walk. This is 1/6 of 240
students.
 150 students ride the bus. This is 3/4 of the
200 students who do not walk.
 25 students ride their bikes. This is 1/2 of
the 50 who do not walk or ride the bus.
Different Ways to find GCF
Text, pp 144
objective:
 use two other ways of finding
the GCF of two numbers.
Example. Find the GCF of 72 and 56.
Method 1. division
72 and 56.
72 ÷ 56 = 1r16
Divide the higher number by the
lower number.
If the remainder is 0, the
lower number is the GCF. If
not divide the divisor by the
56 ÷ 16 = 3r 8 remainder. Continue this
process until the remainder is
16 ÷ 8 = 2r 0 0. The last divisor is the GCF.
Example. Find the GCF of 72 and 56.
Method 2. Subtraction
72 – 56 = 16.
Subtract the lower number
from the higher number.
56 – 16 = 40
Compare the three numbers.
Subtract the lowest from the
next lowest.
Continue the process until
the last two numbers in the
sentence are the same. That
number is the GCF.
40 – 16 = 24
24 – 16 = 8
16 – 8 = 8
Class work
PB, p 144
Homework
PB, p 159