Download Lesson 1-4

Five-Minute Check (over Lesson 1–3)
Then/Now
New Vocabulary
Key Concept: Absolute Value
Example 1: Evaluate an Expression with Absolute Value
Example 2: Solve an Absolute Value Equation
Example 3: No Solution
Example 4: One Solution
Over Lesson 1–3
Which algebraic expression represents the
verbal expression three times the sum of a
number and its square?
A. 3(x2)
C. 3(x + x2)
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B
D. 3 + x + x2
A
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A
B
C
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D
B. 3x + x
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B.
C.
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D.
C
2
Over Lesson 1–3
Which algebraic expression represents the
verbal expression five less than the product of
the cube of a number and –4?
A. 5 – (–4n3)
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B
D. n3 – 5
A
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A
B
C
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D
C. –4n3 + 5
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B.
C.
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D.
C
B. –4n – 5
3
Over Lesson 1–3
Which equation represents the verbal expression
the sum of 23 and twice a number is 65?
A. 23 + 2(65) = 65
0%
B
0%
A
D. 2n + 23 = 65
A
B
C
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D
C. 23 = 2n + 65
A.
B.
C.
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D.
C
B. 23 + n = 65
Over Lesson 1–3
Solve the equation 12f – 4 = 7 + f.
A. 1
B. 0.5
A
B
C
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D
D
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B
0%
A
D. –1
C
C. 0
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Over Lesson 1–3
Solve the equation 10y + 1 = 3(–2y – 5).
A. 2
B. 1
A
B
C
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D
D
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B
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A
D. –1
C
C. 0
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B.
C.
0%
D.
Over Lesson 1–3
0%
0%
0%
A
B
C
D
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D
D.
C
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B
B.
A
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A.
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D.
You solved equations using properties of
equality. (Lesson 1–3)
• Evaluate expressions involving absolute
values.
• Solve absolute value equations.
• absolute value
• empty set
• extraneous solution
Evaluate an Expression with Absolute Value
Replace x with 4.
Multiply 2 and 4 first.
Subtract 8 from 6.
Add.
Answer: 4.7
A. 18.3
B. 1.7
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B
A
0%
A
B
C
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D
D
D. –13.7
C
C. –1.7
A.
B.
C.
0%
D.
Solve an Absolute Value Equation
Case 1 a = b
Case 2
a = –b
y+3 =8
y + 3 = –8
y+3–3 =8–3
y + 3 – 3 = –8 – 3
y = –11
y=5
Check
|y + 3| = 8
?
|5 + 3| = 8
?
|8| = 8
|y + 3| = 8
?
|–11 + 3| = 8
?
|–8| = 8
8=8
8 = 8
Answer: The solutions are 5 and –11.
Thus, the solution set is –11, 5.
What is the solution to |2x + 5| = 15?
A. {5}
B. {–10, 5}
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B
A
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A
B
C
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D
D
D. {–5}
C
C. {–5, 10}
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B.
C.
0%
D.
No Solution
Solve |6 – 4t| + 5 = 0.
|6 – 4t| + 5 = 0
|6 – 4t| = –5
Original equation
Subtract 5 from each side.
This sentence is never true.
Answer: The solution set is .
A.
0%
B
0%
A
D.
A
B
C
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D
D
C.
A.
B.
C.
0%
D.
C
B.
One Solution
Case 1 a = b
8 + y = 2y – 3
8 =y–3
11 = y
One Solution
Check:

Answer:
A.
B.
0%
B
A
0%
A
B
C
0%
D
D
D.
C
C.
A.
B.
C.
0%
D.