M1 Lesson 11

Lesson 11
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
ALGEBRA I
Exit Ticket Sample Solutions
1.
Here is the graphical representation of a set of real numbers:
a.
Describe this set of real numbers in words.
The set of all real numbers less than or equal to two
b.
Describe this set of real numbers in set notation.
{𝒓 real | 𝒓 ≀ 𝟐} (students might use any variable)
c.
Write an equation or an inequality that has the set above as its solution set.
π’˜ – πŸ• ≀ βˆ’πŸ“ (Answers will, of course, vary, students might use any variable.)
2.
Indicate whether each of the following equations is sure to have a solution set of all real numbers. Explain your
answers for each.
d.
πŸ‘(𝒙 + 𝟏) = πŸ‘π’™ + 𝟏
No, the two algebraic expressions are not equivalent.
e.
𝒙 + 𝟐 = 𝟐 +𝒙
Yes, the two expressions are algebraically equivalent by application of the Commutative Property.
πŸ’π’™(𝒙 + 𝟏) = πŸ’π’™ + πŸ’π’™πŸ
f.
Yes, the two expressions are algebraically equivalent by application of the Distributive Property.
g.
πŸ‘π’™(πŸ’π’™)(πŸπ’™) = πŸ•πŸπ’™πŸ‘
No, the two algebraic expressions are not equivalent.
Problem Set Sample Solutions
For each solution set graphed below, (a) describe the solution set in words, (b) describe the solution set in set notation,
and (c) write an equation or an inequality that has the given solution set.
1.
2.
a. the set of all real numbers equal to πŸ“
a. the set of all real numbers equal to
c. answers vary
c. answers vary
b. {πŸ“}
Lesson 11:
Date:
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b.
𝟐
οΏ½πŸ‘οΏ½
𝟐
πŸ‘
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Lesson 11
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
ALGEBRA I
4.
3.
a. the set of all real numbers greater than 𝟏
a. the set of all real numbers less or equal to πŸ“
c. answers vary
c. answers vary
b. {𝒙 real | 𝒙 > 𝟏}
5.
b. {𝒙 real | 𝒙 ≀ πŸ“}
6.
a. the set of all real numbers not equal to 𝟐
a. the set of all real numbers not equal to πŸ’
c. answers vary
c. answers vary
b. {𝒙 real | 𝒙 β‰  𝟐}
7.
b. {𝒙 real | 𝒙 β‰  πŸ’}
8.
a. the null set
a. the set of all real numbers
b. { } or βˆ…
b. {𝒙 real}
c. answers vary
Lesson 11:
Date:
© 2013 Common Core, Inc. Some rights reserved. commoncore.org
c. answers vary
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Lesson 11
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
ALGEBRA I
Fill in the chart below.
SOLUTION SET IN
WORDS
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
SOLUTION SET
IN SET
NOTATION
The set of real numbers
equal to 𝟐
𝒛 = 𝟐
{𝟐}
The set of real numbers
equal to 𝟐 or βˆ’πŸ
π’›πŸ = πŸ’
{βˆ’πŸ, 𝟐}
The set of real numbers not
𝟏
equal to .
πŸ’π’› β‰  𝟐
{𝒛 real | z β‰  1/2}
𝟐
The set of real numbers
equal to πŸ“
𝒛– πŸ‘ = 𝟐
{πŸ“}
The set of real numbers
equal to 1 or -1
π’›πŸ + 𝟏 = 𝟐
{βˆ’πŸ, 𝟏}
The set of real numbers
equal to 0
𝒛 = πŸπ’›
{𝟎}
The set of real numbers
greater than 2
𝒛 > 𝟐
{𝒛 real | 𝒛 > 2}
The null set
𝒛– πŸ” = 𝒛– 𝟐
{ }
The set of real numbers less
than 4
𝒛 – πŸ” < βˆ’πŸ
{𝒛 real | 𝒛 < 4}
The set of real numbers
πŸ’(𝒛 – 𝟏) β‰₯ πŸ’π’› – πŸ’
GRAPH
ℝ
For problems 19–24, answer the following: Are the two expressions algebraically equivalent? If so, state the property (or
properties) displayed. If not, state why (the solution set may suffice as a reason) and change the equation, ever so
slightly, e.g., touch it up, to create an equation whose solution set is all real numbers.
19.
𝒙(πŸ’ – π’™πŸ ) = (βˆ’π’™πŸ + πŸ’)𝒙
Yes, commutative
20.
πŸπ’™
πŸπ’™
=𝟏
No, the solution set is {𝒙 real | 𝒙 β‰  𝟎}. If we changed it to:
numbers.
πŸπ’™
𝟏
= πŸπ’™, it would have a solution set of all real
21. (𝒙 – 𝟏)(𝒙 + 𝟐) + (𝒙 – 𝟏)(𝒙 – πŸ“) = (𝒙 – 𝟏)(πŸπ’™ – πŸ‘)
Yes, distributive.
22.
𝒙
πŸ“
𝒙
+ =
πŸ‘
πŸπ’™
πŸ–
No, the solution set is {𝟎}. It we changed it to:
Lesson 11:
Date:
© 2013 Common Core, Inc. Some rights reserved. commoncore.org
𝒙
πŸ“
𝒙
+ =
πŸ‘
πŸ–π’™
πŸπŸ“
, it would have a solution set of all real numbers.
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Lesson 11
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M1
ALGEBRA I
23. π’™πŸ + πŸπ’™πŸ‘ + πŸ‘π’™πŸ’ = πŸ”π’™πŸ—
No, neither the coefficients nor the exponents are added correctly. One way it could have a solution set of all real
numbers would be:
π’™πŸ + πŸπ’™πŸ‘ + πŸ‘π’™πŸ’ = π’™πŸ (𝟏 + πŸπ’™ + πŸ‘π’™πŸ ).
24. 𝒙
πŸ‘
+ πŸ’π’™ + πŸ’π’™ = 𝒙(𝒙 + 𝟐)
𝟐
Yes, distributive
25. Solve for π’˜:
πŸ”π’˜+𝟏
πŸ“
𝟏
οΏ½π’˜ 𝒓𝒆𝒂𝒍 | π’˜ β‰  𝟏 οΏ½
𝟐
β‰  𝟐. Describe the solution set in set notation.
𝟐
26. Edwina has two sticks, one 𝟐 yards long and the other 𝟐 meters long. She is going to use them, with a third stick of
some positive length, to make a triangle. She has decided to measure the length of the third stick in units of feet.
a.
What is the solution set to the statement: β€œSticks of lengths 𝟐 yards, 𝟐 meters and 𝑳 feet make a triangle”?
Describe the solution set in words, and through a graphical representation.
One meter is equivalent, to two decimal places, to πŸ‘. πŸπŸ– feet. We have that 𝑳 must be a positive length
greater than 𝟎. πŸ“πŸ” feet and less than πŸ”. πŸ“πŸ” feet.
b.
What is the solution set to the statement: β€œSticks of lengths 𝟐 yards, 𝟐 meters and 𝑳 feet make an isosceles
triangle”? Describe the solution set in words and through a graphical representation.
𝑳 = πŸ” feet or 𝑳 = πŸ”. πŸπŸ– feet.
c.
What is the solution set to the statement: β€œSticks of lengths 𝟐 yards, 𝟐 meters and 𝑳 feet make an equilateral
triangle”? Describe the solution set in words, and through a graphical representation.
The solution set is the empty set.
Lesson 11:
Date:
© 2013 Common Core, Inc. Some rights reserved. commoncore.org
Solution Sets for Equations and Inequalities
8/9/13
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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
154