Transcript 1.3 powerpoint

Content Standards
A.CED.1 Create equations and inequalities
in one variable and use them to solve
problems.
Mathematical Practices
3 Construct viable arguments and critique
the reasoning of others.
8 Look for and express regularity in repeated
reasoning.
You used properties of real numbers to
evaluate expressions.
• Translate verbal expressions into algebraic
expressions and equations, and vice versa.
• Solve equations using the properties of
equality.
• open sentence
• equation
• solution
Verbal to Algebraic Expression
A. Write an algebraic expression to represent the
verbal expression 7 less than a number.
Answer:
Verbal to Algebraic Expression
A. Write an algebraic expression to represent the
verbal expression 7 less than a number.
Answer: n – 7
Verbal to Algebraic Expression
B. Write an algebraic expression to represent the
verbal expression the square of a number
decreased by the product of 5 and the number.
Answer:
Verbal to Algebraic Expression
B. Write an algebraic expression to represent the
verbal expression the square of a number
decreased by the product of 5 and the number.
Answer: x2 – 5x
A. Write an algebraic expression to represent the
verbal expression 6 more than a number.
A. 6x
B. x + 6
C. x6
D. x – 6
A. Write an algebraic expression to represent the
verbal expression 6 more than a number.
A. 6x
B. x + 6
C. x6
D. x – 6
B. Write an algebraic expression to represent the
verbal expression 2 less than the cube of a number.
A. x3 – 2
B. 2x3
C. x2 – 2
D. 2 + x3
B. Write an algebraic expression to represent the
verbal expression 2 less than the cube of a number.
A. x3 – 2
B. 2x3
C. x2 – 2
D. 2 + x3
Algebraic to Verbal Sentence
A. Write a verbal sentence to represent 6 = –5 + x.
Answer:
Algebraic to Verbal Sentence
A. Write a verbal sentence to represent 6 = –5 + x.
Answer: Six is equal to –5 plus a number.
Algebraic to Verbal Sentence
B. Write a verbal sentence to represent 7y – 2 = 19.
Answer:
Algebraic to Verbal Sentence
B. Write a verbal sentence to represent 7y – 2 = 19.
Answer: Seven times a number minus 2 is 19.
A. What is a verbal sentence that represents the
equation n – 3 = 7?
A. The difference of a number
and 3 is 7.
B. The sum of a number and
3 is 7.
C. The difference of 3 and a
number is 7.
D. The difference of a number
and 7 is 3.
A. What is a verbal sentence that represents the
equation n – 3 = 7?
A. The difference of a number
and 3 is 7.
B. The sum of a number and
3 is 7.
C. The difference of 3 and a
number is 7.
D. The difference of a number
and 7 is 3.
B. What is a verbal sentence that represents the
equation 5 = 2 + x?
A. Five is equal to the difference
of 2 and a number.
B. Five is equal to twice a
number.
C. Five is equal to the quotient
of 2 and a number.
D. Five is equal to the sum of
2 and a number.
B. What is a verbal sentence that represents the
equation 5 = 2 + x?
A. Five is equal to the difference
of 2 and a number.
B. Five is equal to twice a
number.
C. Five is equal to the quotient
of 2 and a number.
D. Five is equal to the sum of
2 and a number.
Solve One-Step Equations
A. Solve m – 5.48 = 0.02. Check your solution.
m – 5.48 = 0.02
Original equation
m – 5.48 + 5.48 = 0.02 + 5.48
Add 5.48 to each
side.
m = 5.5
Check m – 5.48 = 0.02
?
5.5 – 5.48 = 0.02
0.02 = 0.02 
Answer:
Simplify.
Original equation
Substitute 5.5 for m.
Simplify.
Solve One-Step Equations
Original equation
Simplify.
Solve One-Step Equations
Check
Original equation
?
Substitute 36 for t.

Answer:
Simplify.
A. What is the solution to the equation x + 5 = 3?
A. –8
B. –2
C. 2
D. 8
A. What is the solution to the equation x + 5 = 3?
A. –8
B. –2
C. 2
D. 8
B. What is the solution to the equation
A. 5
B.
C. 15
D. 30
B. What is the solution to the equation
A. 5
B.
C. 15
D. 30
Solve a Multi-Step Equation
Solve 53 = 3(y – 2) – 2(3y – 1).
53 = 3(y – 2) – 2(3y – 1)
Original equation
53 = 3y – 6 – 6y + 2
Apply the Distributive
Property.
53 = –3y – 4
Simplify the right side.
57 = –3y
Add 4 to each side.
–19 = y
Answer:
Divide each side by –3.
Solve a Multi-Step Equation
Solve 53 = 3(y – 2) – 2(3y – 1).
53 = 3(y – 2) – 2(3y – 1)
Original equation
53 = 3y – 6 – 6y + 2
Apply the Distributive
Property.
53 = –3y – 4
Simplify the right side.
57 = –3y
Add 4 to each side.
–19 = y
Divide each side by –3.
Answer: The solution is –19.
What is the solution to 25 = 3(2x + 2) – 5(2x + 1)?
A. –6
B.
C.
D. 6
What is the solution to 25 = 3(2x + 2) – 5(2x + 1)?
A. –6
B.
C.
D. 6
A
B
C
D
Read the Test Item
You are asked to find the value of the expression 4g – 2.
Your first thought might be to find the value of g and then
evaluate the expression using this value. Notice that you
are not required to find the value of g. Instead, you can
use the Subtraction Property of Equality.
Solve the Test Item
Original equation
Subtract 7 from each side.
Simplify.
Answer:
Solve the Test Item
Original equation
Subtract 7 from each side.
Simplify.
Answer: C
If 2x + 6 = –3, what is the value of 2x – 3?
A. 12
B. 6
C. –6
D. –12
If 2x + 6 = –3, what is the value of 2x – 3?
A. 12
B. 6
C. –6
D. –12