A.SSE.1a/A1.ASE.1/FA.ASE.1/IA.ASE.1* Answers...nx + p, where a, b, c, m, n, and p are real numbers, a ≠ 0, and m ≠ 0, is (a + m)x2 + (b + n)x + (c + p). If a + m = 0, the simplified

Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.

Algebra 1 Teacher Guide 7 Common Core Assessment Readiness

A.SSE.1a/A1.ASE.1*/FA.ASE.1*/IA.ASE.1* Answers 1. C 2. D 3. D 4. B, F, G 5. a. 2.25 + 0.25d b. 0.25 represents the amount of feet

the vine grows in a single day. d represents the number of days after the researcher started observing the vine.

c. 0.25d represents the amount of feet the vine has grown after the researcher started observing it.

Rubric a. 1 point b. 0.5 point for correctly interpreting 0.25;

0.5 point for correctly interpreting d c. 1 point for identifying 0.25d as the

amount of feet the vine has grown since the researcher started observing the vine

6. The algebraic expression is 50 − 2d. The −2 represents the number of gallons of water lost each day. The d represents the number of days after the tank is filled. Rubric 1 point for a correct expression; 1 point for a correct interpretation of −2; 1 point for a correct interpretation of d

7. a. s represents a length. s3 represents the volume of a cube. 0.5s represents a radius of a cylinder that is half the width of the cube, and (0.5s)2πh represents the volume of a cylinder with diameter s and height h.

b. The sketch should look like a cube with a cylinder attached.

Rubric a. 0.5 point for each accurate

interpretation b. 2 points for a reasonable sketch

8. 4(3w); 4 represents the area of the algae that originally covered the pond; 3w represents the area of algae that covers the pond after w weeks for every square foot of algae that originally covered the pond. Rubric 2 points for the correct expression; 1 point for the correct interpretation of 4; 1 point for the correct interpretation of 3w



A.SSE.1b/A1.ASE.1*/FA.ASE.1*/IA.ASE.1* Answers 1. C 2. D 3. B 4. B, E 5. G 6. B 7. F 8. C 9. 250 − 5(a − 1) is the price they can

charge per ad for a ads. For one ad, they can charge 250 − 5(1 − 1) = 250 − 5(0) = $250, for two ads, they can charge 250 − 5(2 − 1) = 250 − 5 = $245, and so on, which matches the pattern they found in their market research. Rubric 1 point for interpretation; 1 point for explanation

10. 1.75 represents the initial height of the plant in feet. The expression 1.75(1 + 0.20)d represents the height of the plant in feet after d days.

Rubric 1 point for interpreting 1.75; 1 point for interpreting 1.75(1 + 0.20)d



A.SSE.2/A1.ASE.2*/IA.ASE.2* Answers 1. B 2. D 3. A 4. A, B, D, F 5. B, E 6. 4 binomials

6561− 256y16 = 38 − 28 y16

= 34( )2 − 24 y 8( )2= 34 + 24 y 8( ) 34 − 24 y 8( )= 34 + 24 y 8( ) 32( )2 − 22 y 4( )2⎡

⎣⎢⎤⎦⎥

= 34 + 24 y 8( ) 32 + 22 y 4( ) 32 − 22 y 4( )= 34 + 24 y 8( ) 32 + 22 y 4( ) 3( )2

− 2y 2( )2⎡⎣⎢

⎤⎦⎥

= 34 + 24 y 8( ) 32 + 22 y 4( ) 3 − 2y 2( ) 3+ 2y 2( )

Rubric 1 point for correct number of binomials; 3 points for appropriate work

7. a. 9k4 + 78k2 + 169 = (3k2)2 + 2(3k2)(13) + (13)2

b. (3k2)2 + 2(3k2)(13) + (13)2 = (3k2 + 13)2 Rubric

a. 1 point b. 1 point

8. The length of each deck is the area divided by the width. Al’s deck: (60x + 40) ÷ 4 = 15x + 10 Tim’s deck: (21x + 14) ÷ 7 = 3x + 2 Notice that the terms of the expression for the length of Al’s deck have a common factor. Use the distributive property to rewrite the expression for the length of Al’s deck. 15x + 10 = 5(3x + 2) The length of Tim’s deck is 3x + 2, so the length of Al’s deck is the product of 5 and the length of Tim’s deck. The length of Al’s deck is 5 times the length of Tim’s deck. Rubric 1 point for correct answer; 3 points for appropriate work and explanations

9. a. Yes; (ax + b)2 = (ax)2 + 2(ax)(b) + b2. Since 25x2 = (5x)2 and 81 = 92, a = 5 and b = 9. Since 2(5x)(9) = 90x, 25x2 + 90x + 81 = (5x + 9)2.

b. No; (ay2 + b)2 = (ay2)2 + 2(ay2)(b) + b2. Since 9y4 = (3y2)2 and 49 = 72, the potential values are a = 3 and b = 7. However, 2(3y2)(7) = 42y2 ≠ 30y2, so there are no values of a and b that satisfy 9y4 + 30y2 + 49 = (ay2 + b)2.

Rubric a. 1 point for answer; 1 point for

explanation b. 1 point for answer; 1 point for

explanation



A.SSE.3a/A1.ASE.3*a/IA.ASE.3*a Answers 1. B 2. B, E, F 3. F 4. C 5. E 6. D 7.

n2 − 23n +132 = 0(n −12)(n −11) = 0

n = 11, 12

There are two zeros, 11 and 12. Rubric 0.5 point for each zero; 1 point for a correct factorization

8.

3x2 + 8x − 35 = 0(3x − 7)(x + 5) = 0

x = 73

,− 5

There are two zeros, 73

and −5.

Rubric 0.5 point for each zero; 1 point for a correct factorization

9. When factoring x2 + bx + c, the goal is to find two numbers that add to b and multiply to c. If the two numbers had the same sign, their product, c, would be positive. Since c is negative, the two numbers must have opposite signs. Rubric 3 points for an accurate explanation

10.

−16t2 + 64t = 0−16(t2 − 4t) = 0

t2 − 4t = 0t(t − 4) = 0

t = 0,4 The boulder is on the ground at t = 0 and t = 4. At t = 0, the boulder has not yet been launched. It takes 4 seconds for the boulder to hit the ground after the launch.

Rubric 1 point for answer; 3 points for work

11. The ball passes through the hoop when h(t) = 0.

−16t2 + 4t + 240 = 04t2 − t − 60 = 0

(4t +15)(t − 4) = 0

t = −154

, 4

Since t is the time in seconds after the ball

is thrown, −15

4 does not make sense in

the context of the problem. It takes the ball 4 seconds after it is thrown before it passes through the hoop. Rubric 1 point for the correct answer; 3 points for appropriate work

12. a. If c < 0, then one number is positive and one is negative.

b. If b > 0, then the absolute value of the positive number is greater than the absolute value of the negative number. This means that the absolute value of the negative zero is greater than the absolute value of the positive zero.

c. Since c is still less than 0, the two numbers must have opposite signs. The answer to part a does not change. If b < 0, then the absolute value of the negative number is greater than the absolute value of the positive number. This means that the absolute value of the positive zero is greater than the absolute value of the negative zero. The answer for part b is reversed.

Rubric a. 1 point for the correct answer; 1 point

for a correct explanation b. 1 point for the correct answer; 1 point for

a correct explanation c. 1 point for noting answer a is the same;

1 point for noting answer b is reversed; 1 point for explanation



A.APR.1/A1.AAPR.1*/IA.AAPR.1*1 Answers 1. C 2. B 3. C 4. C, E 5. a. Yes b. No c. No d. No e. Yes 6. 6x2 − 13x − 8

Rubric 2 points

7. No. The sum of two quadratic polynomials ax2 + bx + c and mx2 + nx + p, where a, b, c, m, n, and p are real numbers, a ≠ 0, and m ≠ 0, is (a + m)x2 + (b + n)x + (c + p). If a + m = 0, the simplified form of the sum will be (b + n)x + (c + p), which does not have an x2 term. Rubric 1 point for correct answer; 2 points for reasonable explanation

8. Gross revenue as the product of two factors: (65 + 5x)(30 − 2x)

(65 + 5x)(30 − 2x)= 1950 −130x +150x −10x2

= −10x2 + 20x +1950

Gross revenue in expanded form: −10x2 + 20x + 1950 Rubric 1 point for gross revenue as the product of two factors; 1 point for gross revenue in expanded form; 2 points for appropriate work

9. a. 65t + 1.2 b. 55t + 0.7

c.

(65t +1.2)− (55t + 0.7)= 65t +1.2− 55t − 0.7= 10t + 0.5

The cars are 10t + 0.5 miles apart. d. (65t + 1.2) + (55t + 0.7) = 120t + 1.9 The cars are 120t + 1.9 miles apart.

Rubric a. 1 point b. 1 point c. 1 point for answer, 1 point for work d. 1 point for answer, 1 point for work



A.CED.1/A1.ACE.1*/FA.ACE.1*/IA.ACE.1* Answers 1. B 2. D 3. A 4. D 5. C 6. B, C 7. a. Yes b. No c. Yes d. No 8. 0.7d + 0.25(58) = 58, where d is the

remaining download time in seconds; d ≈ 62 seconds Rubric 1 point for writing a correct equation; 1 point for solving and rounding answer correctly

9. The expression 1.75b − 2 gives the correct price only when b ≥ 6, so Shauna cannot use this function to find the price for fewer than 6 bottles of water. Three bottles of water would cost $5.25, which is more than the $5 she has. Rubric 1 point for explaining the function does not give correct values when b < 6; 1 point for showing that 3 bottles cost more than the money she has

10. a. 2x < x − 3 + 7 → x < 4 x − 3 < 2x + 7 → x > − 10

7 < 2x + x − 3 → x > 3 1

3

b. 3 1

3< x < 4 ; If you graph all three

inequalities on a number line, the

interval of 3 1

3< x < 4 is the only

place where all three inequalities are satisfied.

c. Possible answer: When x = 3 1

2, the

side lengths are 7, 7, and 12

units.

Rubric a. 1 point for correctly writing the three

inequalities; 1 point for solving the inequalities

b. 1 point for finding the correct range; 1 point for supplying an appropriate explanation

c. 1 point for choosing a value in the range found in part b; 1 point for evaluating the side lengths using the value



A.CED.2/A1.ACE.2*/FA.ACE.2*/IA.ACE.2* Answers 1. D 2. A 3. P = 250 + 25d

Rubric 2 points

4. m = 50(0.995)t Rubric 2 points

5. Possible answer:

Rubric 0.5 point for each reasonable scale; 1 point for correct graph

6. h = −16t2 + 96t + 3 Possible graph:

−16t2 + 96t + 3 = 0−16 t2 − 6t( ) + 3 = 0

−16 t2 − 6t + 9( ) + 3 +144 = 0

−16 t − 3( )2 +147 = 0

The maximum height of the ball is 147 feet. Rubric 1 point for correct equation; 0.5 point each for appropriate axes scales given the student’s axis label choices; 1 point for accurate graph; 1 point for correct maximum height; 1 point for accurate work



7. a. d = 10.38t b. Possible graph:

c. The competitor’s distance in meters

d after t seconds is d = 7t + 30. Graphing the competitor’s distance on the same graph as Bolt’s distance results in the following (possible) graph.

Since Bolt reaches 100 meters first, he would win the race.

Rubric a. 2 points b. 0.5 point each for appropriate axes

scales given the student’s axis label choices; 1 point for accurate graph

c. 1 point for correct answer; 1 point for accurate justification



A.CED.4/A1.ACE.4*/FA.ACE.4*/IA.ACE.4* Answers 1. B 2. A, D 3. F 4. A 5. C 6.

P = 8.4Y + 330T +100C − 200IA

PA = 8.4Y + 330T +100C − 200IPA+ 200I − 8.4Y − 330T = 100CPA+ 200I − 8.4Y − 330T

100= C

Rubric 1 point for correct equation; 1 point for accurate work

7. a. P = 8r + 2πr b.

P = 8r +2πrP = r(8 +2π )

P8 +2π

= r

Rubric a. 2 points for correct equation for P in

terms of r b. 1 point for correct equation for r in

terms of P; 1 point for accurate work

8. a.

D = (x2 − x1)2 + (y2 − y1)

2

D2 = (x2 − x1)2 + (y2 − y1)

2

D2 − (x2 − x1)2 = (y2 − y1)

2

± D2 − (x2 − x1)2 = y2 − y1

y1 ± D2 − (x2 − x1)2 = y2

b.

y = −3 ± 102 − [1− (−5)]2

= −3 ± 100 − 62

= −3 ± 100 − 36

= −3 ± 64= −3 ±8= 5 or −11

The slope of the line through (−5, −3) and

(1, 5) is

5 − (−3)1− (−5)

= 86= 4

3. Since the

resulting line has a positive slope, 5 is a possible value of y. The slope of the line through (−5, −3) and

(1, −11) is

−11− (−3)1− (−5)

= −86= − 4

3. Since

the resulting line has a negative slope, −11 is not a possible value of y. The value of y is 5. Rubric

a. 1 point for correct equation solved for y2; 1 point for accurate work

b. 1 point for correct answer; 2 points for correct work and reasoning



A.REI.1/A1.AREI.1*/FA.AREI.1* Answers 1. A 2. D 3. A, B, E, F 4. a. Distributive Property b. Subtraction Property of Equality c. Division Property of Equality

Rubric a. 1 point for correct property b. 1 point for correct property

(award full credit for “Addition Property of Equality”)

c. 1 point for correct property (award full credit for “Multiplication Property of Equality”)

5. a. D = ts + 101 b. Possible answer:

410 = 5s +101410 −101= 5s +101−101

(Subtraction Propertyof Equality)

309 = 5s3095

= 5s5

(Division Propertyof Equality)

61.8 = s

Dennis’s average speed from Hartford to Baltimore is 61.8 miles per hour.

Rubric a. 1 point b. 1 point for setting up correct equation;

2 points for reasonable solution steps with correct justifications 1 point for correct answer of 61.8 miles per hour

6.

−16t2 + 32t + 70 = 22−16t2 + 32t + 70 − 22 = 22− 22

(Subtraction Propertyof Equality)

−16t2 + 32t + 48 = 0−16(t − 3)(t +1) = 0

(t − 3)(t +1) = 0(Division Propertyof Equality)

t − 3 = 0 or t +1= 0(Zero Propertyof Equality)

t − 3 + 3 = 0 + 3(Addition Propertyof Equality)

t = 3

t +1−1= 0 −1(Subtraction Propertyof Equality)

t = −1

Because t represents time, the negative solution does not make sense in this context. It takes the rock 3 seconds before it is 22 feet above the water. Rubric 3 points for reasonable solution steps with correct justifications; 0.5 point for each solution to the equation; 1 point for a correct conclusion based on the context



A.REI.3/A1.AREI.3*/FA.AREI.3* Answers 1. D 2. D 3. B, D 4. a. No b. No c. Yes d. Yes e. No 5.

−4x + 5 ≥ −23−4x + 5 − 5 ≥ −23 − 5

−4x ≥ −28−4x−4

≤ −28−4

x ≤ 7

Rubric 1 point for a correct inequality; 1 point for accurate work

6.

7 −4− 83

x⎛⎝⎜

⎞⎠⎟

−5= 28

7 −4− 83

x⎛⎝⎜

⎞⎠⎟= −5 i 28

7 −4− 83

x⎛⎝⎜

⎞⎠⎟= −140

−4− 83

x = −1407

−4− 83

x = −20

− 83

x = −20 + 4

− 83

x = −16

x = − 38

i −16

x = 6

Rubric 1 point for correct answer; 2 points for accurate work

7. Let t be the time in hours that Gwendolyn spends reading.

130 + 45t = 400130 + 45t −130 = 400 −130

45t = 27045t45

= 27045

t = 6

Gwendolyn needs 6 hours to finish the book. Rubric 1 point for correct answer; 1 point for accurate work

8. P = 2a − 50

2a − 50 = 1302a − 50 + 50 = 130 + 50

2a = 1802a2

= 1802

a = 90

Annika needs to sell 90 prints. A convention that runs from Friday to Sunday lasts three days, so Annika needs to sell 90 ÷ 3 = 30 prints per day. Rubric 1 point for correct original equation; 1 point for correct number of prints; 1 point for accurate work; 1 point for correct number of prints per day



9. a. H ≤ 8r + 16 b.

68 ≤ 8r +1668 −16 ≤ 8r +16 −16

52 ≤ 8r528

≤ 8r8

6.5 ≤ r

The second competitor has to eat at least 6.5 hot dogs and buns per minute to tie or break the record.

c. Possible answer: The second competitor would break the record. His current pace is 8 hot dogs and buns per minute, and he only needs to eat more than 6.5 hot dogs and buns per minute to break the record.

Rubric a. 1 point b. 1 point for correct conclusion of “at

least 6.5”; 1 point for accurate work c. 1 point for correct answer; 1 point for

reasonable justification



A.REI.4a/A1.AREI.4*a/IA.AREI.4*a Answers 1. B 2. B 3. A, C, D 4. a. No b. Yes c. Yes d. Yes e. No 5.

x2 +14x +13 = 0x2 +14x = −13

x2 +14x + 49 = −13 + 49(x + 7)2 = 36

Rubric 1 point for correct form; 2 points for accurate work

6. 2

2

2

2

2

2

4 16 21 124 16 33

3344334 4 4433 16( 2)4 49( 2)4

x xx x

x x

x x

x

x

− − =− =

− =

− + = +

− = +

4− =

Rubric 1 point for correct form; 2 points for accurate work

7. 2

2

2

2

2

3 18 77 23 18 75

6 256 9 25 9( 3) 16

x xx xx x

x xx

+ + =+ = −+ = −

+ + = − ++ = −

The original equation has no real solutions, as it can be rewritten to say the square of a binomial is equal to a negative number.

Rubric 2 points for correctly completing the square; 1 point for correct conclusion

8.

ax2 + bx + c = 0ax2 + bx = −c

x2 + ba

x = − ca

x2 + ba

x + b2

4a2 = − ca+ b2

4a2

x + b2a

⎛⎝⎜

⎞⎠⎟

2

= b2

4a2 − 4ac4a2

x + b2a

= ± b2 − 4ac4a2

x + b2a

= ± b2 − 4ac2a

x = −b ± b2 − 4ac2a

Rubric 4 points

9. a. 2

2

2

2

2

16 64 217 2516 64 192

4 124 4 12 4( 2) 16

t tt tt t

t tt

− + + =− + = −

− =− + = +

− = b. Solve the equation from part a to find

how long the boulder is in the air before it strikes the castle wall.

2( 2) 16

2 162 42 4 or 2 4

6 2

tttt t

t t

− =− = ±− = ±− = − = −

= = −

Exclude −2 as a possible solution because the value of time in this situation cannot be negative. The boulder was in the air for 6 seconds before it struck the wall.



c. Use the fact that distance is the product of rate and time, d = rt, to find the horizontal distance between the catapult and the castle wall.

(95 feet per second)(6 seconds)570 feet

d rtd===

The catapult is 570 feet away from the castle wall.

Rubric a. 1 point for the correct equation in

the correct form; 2 points for accurate work

b. 1 point for accurate work; 1 point for answer

c. 1 point for accurate work; 1 point for answer



A.REI.4b/ A1.AREI.4*b/ IA.AREI.4*b Answers 1. C 2. B 3. A, B, D 4. a. One b. Zero c. Two d. One e. Two 5.

2 11 24 0( 3)( 8) 0

3 or 8

x xx x

x x

− + =− − =

= =

Rubric 1 point for each answer; 1 point for appropriate work

6. Possible answer: (completing the square)

2

2

2

22 5722 121 57 121( 11) 64

11 6411 8 19 or 3

x xx x

x

xx

− = −

− + = − +

− =

− = ±= ± =

Rubric 1 point for each solution; 1 point for appropriate work

7. A quadratic equation has no real solutions when the discriminant of the quadratic formula, b2 − 4ac, is less than zero.

2

2

2

4(5)(12) 0240 0

240

bb

b

− <

− <

<

b < 15.49 or b > −15.49 5x2 + bx + 12 = 0 has no real solutions for values of −15.49 < b < 15.49. Rubric 1 point for correct answer; 2 points for reasonable justification

8. a. 2

2

2

2

2

2

25 3 16 9616 96 22

116

811

6 9 9811 72

( 3)8 8

61( 3)

861

3861

38

5.76 or 0.24

t tt t

t t

t t

t

t

t

t

t

− = − +

− + =

− = −

− + = − +

− = − +

− =

− = ±

= ±

≈

The ball takes about 5.76 seconds to land in the bleacher seats.

Possible answer: 5.76 was chosen

instead of 0.24 because, even though the ball is 25 feet above the ground after 0.24 seconds, it is traveling upward at that time instead of downward.

b. About 461 feet Rubric

a. 1 point for each solution of the quadratic equation; 1 point for correct answer; 1 point for reasonable explanation

b. 1 point



9.Let h be the height of the box. Then h + 3 is the width of the box and 2h − 3 is the length of

the box.

2 2 2

2

2

2

2( )( 3) 2( )(2 3) 2( 3)(2 3) 6702 6 4 6 4 6 12 18 670

10 6 18 67010 6 688 05 3 344 0

h h h h h hh h h h h h h

h hh hh h

+ + − + + − =

+ + − + − + − =

+ − =

+ − =

+ − =

23 3 4(5)( 344)2(5)

3 688910

8 or 8.6

h− ± − −

=

− ±=

= −

The solution h = −8.6 can be rejected since it is not possible to have negative height.

•

8 3 112 8 3 13

+ =− =

The height of the box is 8 inches, the width of the box is 11 inches, and the length of the box is 13 inches.

8 i 11 i 13 = 1144 The volume of the box is 1144 cubic inches.

Rubric 1 point for each correct dimension; 1 point for accurate work; 1 point for correct volume



A.REI.5/A1.AREI.5/FA.AREI.5 Answers 1. A 2. C, D 3.

2 3 3(4 ) 7 3( 7)2 3 12 3 14

14 14

x y x yx y x y

x

− + + = + −− + + = −

= − Rubric 1 point for new equation; 1 point for simplification

4. a. Possible answer:

2x + y − 2(x + 3y ) = 5 − 2(−5)−5y = 15

b. By the addition and multiplication properties of equality, −5y = 15 is equivalent to 2x + y = 5. Therefore,

the system 5 153 5y

x y− =⎧

⎨ + = −⎩ is equivalent

to the given system, and the systems have the same solution.

Rubric a. 1 point for new equation; 1 point

for simplification b. 1 point 5. a.

2 3 2( ) 4 2(3)2 3 2 2 4 6

4 5 10

x y x yx y x y

x y

+ + + = ++ + + = +

+ = b.

?

?

4(5) 5( 2) 10

20 ( 10) 1010 10

+ − =

+ − ==

Rubric a. 1 point for new equation; 1 point for

simplification b. 1 point

6. 0 0

0 0

0 0

0 0

0 0 0 0

0 0 0 0

0 0( ) ( )

Px Qy RSx Ty U

vSx vTy vUR vSx vTy R vU

Px Qy vSx vTy R vUPx vSx Qy vTy R vUP vS x Q vT y R vU

+ =⎧⎨ + =⎩

+ =+ + = +

+ + + = ++ + + = ++ + + = +

Since Sx0 + Ty0 = U is equivalent to (P + vS)x0 + (Q + vT)y0 = R + vU, both general equations have (x0, y0) as a solution. Since Px + Qy = R also has this solution, the system

( ) ( )Px Qy R

P vS x Q vT y R vU+ =⎧

⎨ + + + = +⎩

is equivalent to

.Px Qy RSx Ty U

+ =⎧⎨ + =⎩

Rubric 5 points for a logically sound argument

7. 4(3 2 ) 3(2 3 ) 4(5) 3(12)

12 8 6 9 20 366 17 16

x y x yx y x y

x y

+ − − = −+ − + = −

+ = − The new system of equations is

3 2 5.

6 17 16x yx y

+ =⎧⎨ + = −⎩

6(3)+17(−2) =?−16

18 − 34 =?−16

−16 = −16

Thus, (3, −2) is a solution to the system

of equations 3 2 5

.6 17 16x yx y

+ =⎧⎨ + = −⎩

Rubric 2 points for finding 6x + 17y = −16; 1 point for accurate intermediate calculations; 2 points for showing that (3, −2) is a solution to the new system of equations



8. a. 4 3 3( ) 1 3( 5)

7 14x y x y

x+ + − = + −

= −

An equivalent system is 7 14

.5

xx y

= −⎧⎨ − = −⎩

b. 4 3 4( ) 1 4( 5)

7 21x y x y

y+ − − = − −

=

An equivalent system is 7 21

.5

yx y

=⎧⎨ − = −⎩

c. Possible answer: I can use the system from part b to write a new system by replacing the first equation with the sum of the first equation and 7 times the second equation: 7 7( ) 21 7( 5)

7 14y x y

x+ − = + −

= −

Replacing the equation 7y = 21 in the

system 7 21

5y

x y=⎧

⎨ − = −⎩ with the

equivalent equation 7x = −14 yields

the other system, 7 14

.5

xx y

= −⎧⎨ − = −⎩

Therefore, the systems are equivalent. Rubric

a. 1 point for new equation; 1 point for writing the system

b. 1 point for new equation; 1 point for writing the system

c. 2 point



A.REI.6/A1.AREI.6*/FA.AREI.6* Answers 1. D 2. A, B 3. a. One b. Infinitely many c. One d. One e. Zero 4. Possible answer:

5 2 42(5 2 ) 2(4)10 4 8

10 4 8(3 4 6)7 14

2

5(2) 2 410 2 4

2 63

x yx yx y

x yx y

xx

yyyy

− =− =− =

− =− − = −

==

− =− =− = −

=

The solution to the system is (2, 3). Rubric 1 point for a correct answer; 2 points for accurate work

5. Possible graph:

The exact solution to the system is

15 13, 4 .19 19

⎛ ⎞−⎜ ⎟⎝ ⎠

Rubric 0.5 point for each correctly graphed line; 1 point for a reasonable estimate of the

exact solution 15 13, 419 19

⎛ ⎞−⎜ ⎟⎝ ⎠

6. a. Possible answer: Let x represent the number of nickels Shawntae has and let y represent the number of dimes Shawntae has.

21

0.05 0.1 1.7x y

x y+ =⎧

⎨ + =⎩

b. Possible answer:

2121

0.05 0.1(21 ) 1.70.05 2.1 0.1 1.7

0.05 0.48

8 2121 813

x yy x

x xx x

xx

yyy

+ == −

+ − =+ − =

− = −=

+ == −=



c. Shawntae has 8 nickels and 13 dimes. Rubric

a. 2 points b. 1 point for solution;

1 point for accurate work c. 1 point 7. a. Possible answer: The variable p

represents the number of prints Chris sold. Chris can only sell a whole number of prints, so a decimal value for p does not make sense.

b. Madeline forgot to distribute the 2 to the p term when simplifying 2(30 ) 5 93.p p− + =

c. 3030

2 5 932(30 ) 5 9360 2 5 93

60 3 933 33

11

11 3030 1119

s ps p

s pp pp p

ppp

sss

+ == −

+ =− + =− + =

+ ===

+ == −=

Chris sold 11 color prints and 19 sketches.

Rubric a. 2 points for a reasonable explanation b. 1 point for correctly identifying

the mistake c. 1 point for a correct answer; 1 point for

accurate work



A.REI.10/A1.AREI.10*/FA.AREI.10 Answers 1. C 2. D 3. C, E, F 4. Yes. The graph of an equation is the set

of all its solutions plotted in the coordinate plane. The graph of this equation would pass through the point (13, 9), so (13, 9) must be a solution of the equation. Rubric 1 point for answer of “Yes”; 1 point for reasonable justification

5. a. Bryce mixed up the line representing −4x + y = −2 with the line representing 2x + y = 2.

b. The graphs of y = 3x − 3 and 2x + y = 2 intersect at the point (1, 0).

c. (0, −2) and (2, 6) Rubric

a. 1 point b. 1 point c. 1 point for each correct intersection

6. a. Possible graph:

b. Possible answer: Yes. The graph

of an equation is the set of all its solutions plotted in the coordinate plane. Since the graph passes through the point (2, 14), it is a solution of Maryse’s equation.

c. Possible answer: No. The graph of an equation is the set of all its solutions plotted in the coordinate plane. Since the graph does not pass through the point (4, 40), it is not a solution of Maryse’s equation.

Rubric a. 2 point for graph b. 0.5 point for answer of “Yes”; 1 point

for reasonable justification c. 0.5 point for answer of “No”; 1 point for

reasonable justification



A.REI.11/A1.AREI.11*/FA.AREI.11*/IA.AREI.11* Answers 1. C 2.

Possible answer: The solutions of the

equation 2x = 4x are approximately x = 0.3 and x = 4. Rubric 0.5 point for each graph; 0.5 point for each reasonable solution estimate

3. a.

b. The approximate solution is x = 6.5. My solution is the best approximation because the difference between 4x + 8 and 7x − 11 is the smallest at x = 6.5.

Rubric a. 2 points for correctly completed table b. 1 point for approximate solution;

1 point for justification

4. Possible answer: Let (a, b) be any point where the graphs of y = f(x) and y = g(x) intersect. This means that f(a) = b and g(a) = b. Since b = b, it follows that f(a) = g(a) by the transitive property of equality. Thus, x = a is a solution of the equation f(x) = g(x), and since (a, b) is any point where the graphs of y = f(x) and y = g(x) intersect, this is true for all such points of intersection. Rubric 3 points for a reasonable answer

5. a. c(t) = 1200(1.015)t k(t) = 1250(1.009)t b.

Year Chris’s Account

Karla’s Account

3 $1254.81 $1284.05 4 $1273.64 $1295.61 5 $1292.74 $1307.27 6 $1312.13 $1319.04 7 $1331.81 $1330.91 8 $1351.79 $1342.89

c. The best approximation of the intersection of the two graphs is at t = 7, because that is the value for which the two accounts are closest in value.

Rubric a. 0.5 point for each correct equation b. 1 point for a correctly completed table c. 1 point for correctly identifying the

correct row; 1 point for explanation

x 4x + 8 7x − 11 5 28 24 5.5 30 27.5 6 32 31 6.5 34 34.5 7 36 38 7.5 38 41.5



6. a. g(t) = 5t + 435, h(t) = 150t b. Possible graph:

c. Possible answer: Yes. The two graphs

intersect at (3, 450), so the hare passes the tortoise after 3 minutes, when they have both traveled 450 meters.

Rubric a. 0.5 point for each correct equation b. 0.5 point for each correct graph c. 1 point for answer of “Yes”;

1 point for reasonable distance and time estimate; 1 point for reasonable justification



A.REI.12/A1.AREI.12*/FA.AREI.12* Answers 1. C 2. B 3. A, D, F 4.

Rubric 1 point for boundary line; 1 point for half-plane

5.

Rubric 0.5 point for each correct boundary line; 0.5 point for each correct half-plane

6.

The system has no solution. Rubric 0.5 point for each correct boundary line; 0.5 point for each correct half-plane; 1 point for description of solution set

7. a.

a + c ≤ 4015a + 8c ≥ 400⎧⎨⎩

b.

c. The bus can hold 20 adults and

15 children, and the company makes a profit because (20, 15) is a solution for both inequalities.

Rubric a. 0.5 point for each inequality b. 0.5 point for each correct line; 0.5

point for each correct half-plane. c. 1 point for answer; 1 point for

explanation

Documents

A.SSE.1a/A1.ASE.1*/FA.ASE.1*/IA.ASE.1* Answers...nx + p, where a, b, c, m, n, and p are real numbers, a ≠ 0, and m ≠ 0, is (a + m)x2 + (b + n)x + (c + p). If a + m = 0, the simplified

A.SSE.1a/A1.ASE.1/FA.ASE.1/IA.ASE.1* Answers...nx + p, where a, b, c, m, n, and p are real numbers, a ≠ 0, and m ≠ 0, is (a + m)x2 + (b + n)x + (c + p). If a + m = 0, the simplified