Download Unit 1-4

Five-Minute Check (over Lesson 1–3)
CCSS
Then/Now
New Vocabulary
Key Concept: Absolute Value
Example 1: Evaluate an Expression with Absolute Value
Example 2: Real-World Example: 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?
Which algebraic expression represents the
verbal expression five less than the product of
the cube of a number and –4?
Which equation represents the verbal expression
the sum of 23 and twice a number is 65?
Solve the equation 12f – 4 = 7 + f.
Solve the equation 10y + 1 = 3(–2y – 5).
Over Lesson 1–3
Which algebraic expression represents the
verbal expression three times the sum of a
number and its square?
A. 3(x2)
B. 3x + x2
C. 3(x + x2)
D. 3 + x + x2
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)
B. –4n3 – 5
C. –4n3 + 5
D. n3 – 5
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
B. 23 + n = 65
C. 23 = 2n + 65
D. 23 + 2n = 65
Over Lesson 1–3
Solve the equation 12f – 4 = 7 + f.
A. 1
B. 0.5
C. 0
D. –1
Over Lesson 1–3
Solve the equation 10y + 1 = 3(–2y – 5).
A. 2
B. 1
C. 0
D. –1
Content Standards
A.SSE.1.b Interpret complicated expressions
by viewing one or more of their parts as a
single entity.
A.CED.1 Create equations and inequalities in
one variable and use them to solve problems.
Mathematical Practices
6 Attend to precision.
You solved equations using properties of
equality.
• Evaluate expressions involving absolute
values.
• Solve absolute value equations.
• absolute value
• empty set
• Ø{ }
• constraint
• 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
C. –1.7
D. –13.7
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}
C. {–5, 10}
D. {–5}
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.
B.
C.
D.
One Solution
Case 1 a = b
8 + y = 2y – 3
8 =y–3
11 = y
One Solution
Check:

Answer:
A.
B.
C.
D.