Download REI Reasoning with Equations and Inequalities Cluster

Mathematics I Resources for EOC Remediation
REI – Reasoning with Equations and Inequalities Cluster:
HSA-REI.A.1
HSA-REI.B.3
HSA-REI.B.4
HSA-REI.C.6
The information in this document is intended to demonstrate the depth and rigor of
the Nevada Academic Content Standards. The items are not to be interpreted as
indicative of items on the EOC exam. These are a collection of standard-based items
for students and only include those standards selected for the Math I EOC
examination.
Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
HSA-REI.A.1 Explain each step in solving a simple equation as following from the equality of
numbers asserted at the previous step, starting from the assumption that the original equation
has a solution. Construct a viable argument to justify a solution method.
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1. Solve the equation below. Justify each step. 2(𝑥 + 1) = 5 𝑥 + 3
Answer:
2( x  1) 
1
x3
5
1
x3
5
10x 10  x 15
2x  2 
Original Equation
Distributive Property
Multiplication Property of Equality
9x  10  15
Subtraction Property of Equality
9x  5
5
x
9
Subtraction Property of Equality
Division Property of Equality
2. Solve the equation below showing all steps.
Answer:
1
2
 x  5  1  3
4
1
2
 x  5  4
4
 x  5
2
 16
x  5  4
x  1 or  9
1
4
(𝑥 + 5)2 − 1 = 3
Original Equation
Addition Property of Equality
Multiplication Property of Equality
Square Root Property
Subtraction Property of Equality
3. Solve the equation 𝑘𝑥 = 𝑚𝑦 − 𝑚𝑥 for 𝑥. Justify all steps.
Answer:
𝑘𝑥 = 𝑚𝑦 − 𝑚𝑥
Original equation
𝑘𝑥 + 𝑚𝑥 = 𝑚𝑦
Addition Property of Equality
𝑥(𝑘 + 𝑚) = 𝑚𝑦
𝑚𝑦
𝑥=
𝑘+𝑚
Distribution Property
Division Property of Equality
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Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
4. Jack solved an equation using the following steps:
Step 1:
Step 2:
Step 3:
9(𝑥 + 1) = 14
9𝑥 + 1 = 14
9𝑥 = 13
13
𝑥=
9
Part A:
Is Jack’s solution correct?
Part B:
If not, find and describe the error.
Part C:
If Jack is incorrect, then correctly solve the equation and justify each step.
Answer:
Part A: No, Part B: Error using the Distributive Property on Step 1 because Jack did
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not distribute to both terms in the parenthesis. Part C: 𝑥 = 9
5. Charlie was solving a problem assigned to her in class. The teacher asked her to create a list of
steps, in order, that will solve the following equation 3( x  1)2  108 .
Choose from the following steps:
A. Add 1 to both sides
B. Divide both sides by 3
C. Multiply both sides by 3
D. Subtract 1 from both sides
E. Square both sides
F. Take the square root of both sides
Step 1: ________
Answer:
Step 2: ________
Step 3: ________
Step 1: B, Step 2: F, Step 3: D
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REI Reasoning with Equations and Inequalities Cluster
6. Daniel is reviewing for the unit exam so he looked back on his notes for solving quadratic
equations by factoring and found the example below:
2x2 – 11x = -15
STEP 1: 2x2 – 11x + 15 = 0
STEP 2: (x – 3)(2x – 5) = 0
STEP 3: x – 3 = 0 or 2x – 5 = 0
STEP 4: x = 3 or 𝑥 =
5
2
Help Daniel in reviewing by explaining each step to him.
Answer:
STEP 1: 2x2 – 11x + 15 = 0
STEP 2: (x – 3)(2x – 5) = 0
STEP 3: x – 3 = 0 or 2x – 5 = 0
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STEP 4: x = 3 or 𝑥 = 2
Write in standard form
Factor
Zero Product Property
Solve for x
7. Identify the missing reasons in the two-column table below.
8(𝑡 + 2) − 3(𝑡 − 4) = 6(𝑡 − 7) + 8
8𝑡 + 16 − 3𝑡 + 12 = 6𝑡 − 42 + 8
8𝑡 − 3𝑡 + 16 + 12 = 6𝑡 − 42 + 8
5𝑡 + 28 = 6𝑡 − 34
5𝑡 − 5𝑡 + 28 = 6𝑡 − 5𝑡 − 34
28 = 𝑡 − 34
28 + 34 = 𝑡 − 34 + 34
62 = 𝑡
Answer:
Given
a.
Commutative property
b.
c.
Simplify
d.
Simplify
a. Distributive property, b. Simplify,
c. Subtraction Property of Equality, d. Addition Property of Equality
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Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
HSA-REI.B.3 Solve linear equations and inequalities in one variable, including equations with
coefficients represented by letters.
1. Fill in a number for each box to create an equation that is true for all values of 𝑥.
2(4𝑥 + 3)(3𝑥 + 5) =
𝑥 2 + 58𝑥 +
Answer: 24𝑥 2 + 58𝑥 + 30
2. Solve and graph the inequality for 𝑦.
Answer:
1 < 13 − 2𝑦
𝑦<6
3. Solve the equation for 𝑥, showing all steps:
6𝑥 + 𝐶𝑥 = 11
Answer:
6𝑥 + 𝐶𝑥 = 11
𝑥(6 + 𝐶) = 11
11
𝑥=
6+𝐶
4. The point (4, 7) is on the graph of 𝑦 = 𝑥 2 + 𝑐. What is the value of 𝑐?
Answer: c  9
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REI Reasoning with Equations and Inequalities Cluster
5. Write a number in each box to make an equation that has an infinite number of solutions.
2(5𝑛 − 3) − 3𝑛 =
𝑛−
Answer: 7n  6
6. Part 1:
Part 2:
Answer:
Solve for x, given that a is a negative integer. 𝑎𝑥 + 8 ≥ 16
Give one value of a, and one value of x that satisfy the inequality above.
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𝑥≤𝑎
Sample answer: a = -2 and
x = -5
HSA-REI.B.4 Solve quadratic equations in one variable.
HSA-REI.B.4a Use the method of completing the square to transform any quadratic equation
in x into an equation of the form (x - p)2 = q that has the same solutions. Derive the quadratic
formula from this form.
HSA-REI.B.4b Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots,
completing the square, the quadratic formula and factoring, as appropriate to the initial form of
the equation. Recognize when the quadratic formula gives complex solutions and write them
as a ± bi for real numbers a and b.
1. The legs of a right triangle are represented by the expression below:
Leg1  x  4
Leg 2  x  6
The area of the triangle is 28 square units. Find the lengths of the legs of the triangle.
Answer: Leg1  14, Leg2  4
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REI Reasoning with Equations and Inequalities Cluster
2. Solve 27𝑥 2 − 48 = 0 for 𝑥.
Answer:
4
𝑥 = ±3
3. Solve −16𝑡 2 − 48𝑡 − 32 = 0 for 𝑡.
Answer:
𝑡 = −1, −2
4. Solve the following quadratic equations by using two different methods. Show your work and
compare the methods.
( x  8)2  36
Part A:
Part B:
Answer:
( x  5)2  40
Part A: Work may vary, 𝑥 = 2, 14
Part B: Work may vary, 𝑥 = 5 ± 2√10
5. For each equation below, identify the value that could be added to each side of the equation in
order to complete the square.
Part A:
Part B:
Part C:
x2  6 x  1
x 2  11x  8
x( x  8)  4
Part D:
x 2  8 x  10  0
Answer:
6 2
11 2
Part A: (− 2) = 9, Part B: ( 2 ) =
121
4
8 2
8 2
, Part C: (− 2) = 16, Part D: (2) = 16
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Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
6. What are the roots of the following equations?
Part A:
Part B:
Part C:
Answer:
x 2  8 x  15  0
x 2  16  0
3x 2  15 x  30  12
Part A: 𝑥 = −3, −5, Part B: 𝑥 = −4, 4, Part C: 𝑥 = −7, 2
7. A shoebox with a square face and a length of 12 inches has a volume of 243 in 3. Each side of the
square face can be represented by the expression x  4.
Part A: How many inches wide and long is the face of the shoebox?
Part B: What are the possible values of x? Why is only one of these values possible?
12
x-4
x-4
Answer: Part A: The length and width of the face of the shoebox is 4.5 inches.
Part B: The possible values are 8.5 and -0.5. Only 8.5 is viable as we cannot have negative
lengths.
HSA-REI.C.6 Solve systems of linear equations exactly and approximately (e.g., with graphs),
focusing on pairs of linear equations in two variables.
1. Consider this system of equations
2 x  5 y  10
6 x  ny  5
What would “n” need to equal for this system to have no solution?
Answer: If n  15 , there would be no solution.
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Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
2. Gabby is at the bottom of a sand dune and Thomas is at the top of the sand dune 1,400 feet
from Gabby. They begin to walk toward each other. Every three seconds, Gabby walks upward
six feet, but then slides downward two feet. Every four seconds, Thomas walks downward ten
feet and then slides downward an additional five feet.
A. Write the functions as time versus the distance traveled for Gabby and Thomas.
B. Graph each function on the same coordinate plane.
C. After how many minutes will they meet? Round to the nearest tenth.
Answer:
A. f1(x) =
4
5
x f2(x) = 1400 – x
3
4
B.
C. They will meet after 275.4 seconds which
equals 4.6 minutes.
3. Mr. Thompson weighs 284 pounds and wants to go on a diet. Mrs. Thompson weighs 110
pounds and wants to gain weight. On March 1st, Mr. Thompson starts a diet to lose weight and
Mrs. Thompson starts a “SEEFOOD” diet to gain weight. Mr. Thompson loses three pounds per
week and Mrs. Thompson gains 1½ pounds per week.
6. Write and graph the function of weight versus time for Mr. and Mrs. Thompson on the
same coordinate plane.
7. During what week will they weigh the same (round the number to the nearest tenth)?
To the nearest pound, determine that weight.
8. The ideal weight for Mr. Thompson’s height is 172 pounds and 140 pounds for Mrs.
Thompson. In what week will each achieve their ideal weight?
Answer:
A.
f1(x) = 284 – 3x, f2(x) = 110 + 1.5x
B. They will weigh the same after 38.7 weeks.
At this time they will both weigh 168 pounds.
C. Mr. Thompson will reach his ideal weight of
172 pounds after 37.3 weeks.
Mrs. Thompson will reach her ideal weight of
140 pounds after 20 weeks.
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REI Reasoning with Equations and Inequalities Cluster
4. Mary lives 660 miles north of Isabelle. They are good friends and are going to get together for
the weekend. Isabelle starts from home and travels at a constant rate of 45 miles per hour.
Mary starts from her house and travels at 65 miles per hour.
A. Write an equation modeling the distance and time traveled by each person.
B. Draw a graph showing each of the models.
C. How long will it take for the two friends to meet?
D. How far did each person travel?
E. Describe why this is important in real life.
Answer:
A. Mary: f (t )  660  65 x , Isabelle: f (t )  45 x
B. See Graph
45 x  660  65 x
C. 110 x  660
6 hours
D.
x6
f (6)  45(6)
f (6)  270
Isabelle will travel 270 miles. Mary will travel 660-270 or 390 miles.
E. They will want to make sure that there is some place to meet at that point. Answers will
vary.
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Initial Draft: Jan. 2017
REI Reasoning with Equations and Inequalities Cluster
5. Lynn was selling raffle tickets at the school carnival. At the end of the night, she closed her
register and noticed a dollar lying on the floor next to it. She said, “I wonder whether the dollar
belongs inside the register or not.” The raffle tickets were 1 ticket for $10 or 2 tickets for $18.
When Lynn looked in her register, she found $497 and 53 ticket stubs. Does the dollar belong in
her register or not? Explain your reasoning mathematically.
Answer:
Yes, there had to be an even number of “$9 tickets” sold and without including the
dollar she found, it would mean there was an odd number of “$9 tickets” sold.
6. Part A:
The point (−2, 3) is a solution of which of the following systems of equations?
Select ALL that apply.
Part B:
Answer:
A.
6𝑥 + 3𝑦 = −3
1
𝑥 + 9𝑦 = 26
2
B.
𝑥 − 4𝑦 = −8
−𝑥 + 3𝑦 = −9
C.
7𝑥 + 3𝑦 = −5
𝑥+𝑦 =5
D.
4𝑥 + 𝑦 = −5
−5𝑥 + 𝑦 = 13
E.
2𝑥 − 𝑦 = −7
2𝑦 = 4𝑥 + 14
F.
𝑥 − 4𝑦 = 10
1
𝑥 − 𝑦 = −2
3
For which system above is (−2, 3) is a unique solution? Select ALL that apply.
Part A: A, D, E, Part B: A, D
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REI Reasoning with Equations and Inequalities Cluster
7. Consider the system below.
{
3𝑥 − 𝑦 = −10
5𝑥 + 11𝑦 = 34
Part A: Solve the system. Show all work.
Part B: Explain which method you used to solve the system and why.
Part C: Determine if the lines are intersecting, parallel, or coincide. Explain your answer.
Answer: Part A:
Substitution Method
5x + 11(3x + 10) = 34
5x + 33x + 110 = 34
38x = -76
x = -2
3(-2) - y = -10
-6 - y = -10
- y = -4
y=4
Elimination Method
11(3x - y) = 11(-10)
33x - 11y = -110
33x - 11y = -110
+ 5x + 11y = 34
38x
= -76
x = -2
3(-2) - y = -10
y=4
Solution: (-2, 4)
Part B: Possible solutions  Substitution Method: I used the substitution method because I noticed it was easy
to solve for y and substitute the value in the other equation.
 Elimination Method: I used the elimination method because I noticed it was easy
to multiply the first equation by 11 to create a zero pair that would eliminate.
Part C: The lines are intersecting because there is one solution.
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Initial Draft: Jan. 2017