Algebra 1 unit 1.5
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UNIT 1.5 ADDING AND SUBTRACTINGUNIT 1.5 ADDING AND SUBTRACTING
REAL NUMBERSREAL NUMBERS
Warm UpSimplify.
1.
3 2. –4
Write an improper fraction to represent eachmixed number.
3. 423
143
4. 767
557
Write a mixed number to represent each improper fraction.
5. 125 2
25
6. 249
223
|–3| –|4|
Add real numbers.
Subtract real numbers.
Objectives
Vocabularyabsolute valueoppositesadditive inverse
All the numbers on a number line are called realnumbers. You can use a number line to modeladdition and subtraction of real numbers.
Addition
To model addition of a positive number, move right. To model addition of a negative number move left.
Subtraction
To model subtraction of a positive number, move left. To model subtraction of a negative number move right.
Example 1A: Adding and Subtracting Numberson a Number line
Add or subtract using a number line.
Start at 0. Move left to –4.
11 10 9 8 7 6 5 4 3 2 1 0
+ (–7)
–4+ (–7) = –11
To add –7, move left 7 units.
–4
–4 + (–7)
Example 1B: Adding and Subtracting Numberson a Number line
Add or subtract using a number line.
Start at 0. Move right to 3.
To subtract –6, move right 6 units.
-3 -2 -1 0 1 2 3 4 5 6 7 8 9
+ 3
3 – (–6) = 9
3 – (–6)
–6
Add or subtract using a number line.
–3 + 7
Check It Out! Example 1a
Start at 0. Move left to –3.
To add 7, move right 7 units.
-3 -2 -1 0 1 2 3 4 5 6 7 8 9
–3
+7
–3 + 7 = 4
Check It Out! Example 1b
Add or subtract using a number line.
–3 – 7 Start at 0. Move left to –3.
To subtract 7 move left 7 units.
–3–7
11 10 9 8 7 6 5 4 3 2 1 0
–3 – 7 = –10
Check It Out! Example 1c
Add or subtract using a number line.
–5 – (–6.5) Start at 0. Move left to –5.To subtract negative 6.5 move right 6.5 units.
8 7 6 5 4 3 2 1 0
–5
–5 – (–6.5) = 1.5
1 2
– (–6.5)
The absolute value of a number is the distance from zero on a number line. The absolute value of 5 is written as |5|.
5 units 5 units
210123456 6543- - - - - -
|5| = 5|–5| = 5
Example 2A: Adding Real Numbers
Add.
Use the sign of the number with the greater absolute value.
The sum is negative.
When the signs of numbers are different, find the difference of the
absolute values:
Example 2B: Adding Real Numbers
Add.
y + (–2) for y = –6
y + (–2) = (–6) + (–2)
(–6) + (–2)
First substitute –6 for y.
When the signs are the same, find the sum of the absolute values: 6 + 2 = 8.
–8 Both numbers are negative, so the sum is negative.
Add.
–5 + (–7)
Check It Out! Example 2a
When the signs are the same, find the sum of the absolute values.
Both numbers are negative, so the sum is negative.
–5 + (–7) = 5 + 7
5 + 7 = 12
–12
Check It Out! Example 2b
Add.
–13.5 + (–22.3)
When the signs are the same, find the sum of the absolute values.
–13.5 + (–22.3)
–35.8 Both numbers are negative so,the sum is negative.
13.5 + 22.3
Check It Out! Example 2c
Add.
x + (–68) for x = 52 First substitute 52 for x.
x + (–68) = 52 + (–68)
68 – 52
When the signs of the numbers are different, find the difference of the absolute values.
–16Use the sign of the number with the greater absolute value. The sum is negative.
Two numbers are opposites if their sum is 0. A number and its opposite are on opposite sides of zero on a number line, but are the same distance from zero. They have the same absolute value.
A number and its opposite are additive inverses.To subtract signed numbers, you can use additiveinverses.
11 – 6 = 5 11 + (–6) = 5
Additive inverses
Subtracting 6 is the sameas adding the inverse of 6.
Subtracting a number is the same as adding theopposite of the number.
Subtract.
–6.7 – 4.1
–6.7 – 4.1 = –6.7 + (–4.1) To subtract 4.1, add –4.1.
When the signs of the numbersare the same, find the sum of theabsolute values: 6.7 + 4.1 = 10.8.
= –10.8 Both numbers are negative, so the sum is negative.
Example 3A: Subtracting Real Numbers
Subtract.
5 – (–4)
5 − (–4) = 5 + 4
9
To subtract –4 add 4.
Find the sum of the absolute values.
Example 3B: Subtracting Real Numbers
Subtract.
Example 3C: Subtracting Real Numbers
First substitute for z.
To subtract , add .
Rewrite with a denominator of 10.
Example 3C Continued
Write the answer in the simplest form. Both numbers are negative, so the sum is negative.
When the signs of the numbers arethe same, find the sum of the absolute values: .
Subtract.
13 – 21
Check It Out! Example 3a
13 – 21 To subtract 21 add –21.
When the signs of the numbers are different, find the difference of the absolute values: 21 – 13 = 8.
Use the sign of the number with the greater absolute value.
–8
= 13 + (–21)
Check It Out! Example 3b
Subtract.
Both numbers are positive so, the sum is positive.
To subtract add .–3 12 3 1
2
When the signs of the numbers are the same, find the sum of the absolute values: = 4.3 1
212
+
4
x – (–12) for x = –14
Check It Out! Example 3c
Subtract.
x – (–12) = –14 – (–12) First substitute –14 for x.
–14 + (12) To subtract –12, add 12.
When the signs of the numbers are different, find the difference of the absolute values: 14 – 12 = 2.
Use the sign of the number with the greater absolute value.
–2
Example 4: Oceanography ApplicationAn iceberg extends 75 feet above the sea. The bottom of the iceberg is at an elevation of –247 feet. What is the height of the iceberg?Find the difference in the elevations of the top of the iceberg andthe bottom of the iceberg.
elevation at top of iceberg
75
Minus elevation at bottomof iceberg
–247
75 – (–247)
75 – (–247) = 75 + 247
= 322The height of the iceberg is 322 feet.
To subtract –247, add 247.Find the sum of the absolute values.
–
Check It Out! Example 4What if…? The tallest known iceberg in the North Atlantic rose 550 feet above the oceans surface. How many feet would it be from the top of the tallest iceberg to the wreckage of the Titanic, which is at an elevation of –12,468 feet?
elevation at top of iceberg
550
Minus elevation of the Titanic –12,468
–
550 – (–12,468)550 – (–12,468) = 550 + 12,468
Distance from the iceberg to the Titanic is 13,018 feet.
To subtract –12,468, add 12,468.
Find the sum of the absolute values.= 13,018
Add or subtract using a number line.
1. –2 + 9 7 2. –5 – (–3) –2
Add or subtract.3. –23 + 42 19 4. 4.5 – (–3.7) 8.2
5.
Lesson Quiz
6. The temperature at 6:00 A.M. was –23°F.At 3:00 P.M. it was 18°F. Find the differencein the temperatures. 41°F
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