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Coordinate Algebra Practice EOCT Answers Unit 2

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Page 1: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

Coordinate AlgebraPractice

EOCT AnswersUnit 2

Page 2: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#1 Unit 2Which equation shows ax – w = 3 solved for w ?

–axax – w = 3

–w = 3 – ax–ax

=–1–w 3 – ax

–1w = –3 + axw = ax – 3

A. w = ax – 3B. w = ax + 3C. w = 3 – axD. w = 3 + ax

Page 3: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#2 Unit 2Which equation is equivalent to

118

3

4

7

xx?

A. 17x = 88

B. 11x = 88

C. 4x = 44

D. 2x = 44

Least Common Denominator

8

7 311

4 8

x x

8 8

1 1

7 3 11

4 8

x x

56 24 88

4 8

x x

14x – 3x = 88

11x = 88

Page 4: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#3 Unit 2Which equation shows 4n = 2(t – 3) solved for t ?

4n = 2(t – 3)

2n = t – 3

24n 2(t – 3)

2=

+3+32n + 3 = t

4n = 2(t – 3)

4n = 2t – 6+6+6

4n + 6 = 2t

2n + 3 = t

Method #1 Method #2

4n + 6 2t22 =

Page 5: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#4 Unit 2Which equation shows 6(x + 4) = 2(y + 5) solved for y ?

6(x + 4) = 2(y + 5)

–106x + 24 = 2y + 10

–106x + 14 = 2y

3x + 7 = y

6x + 14 2y22

=

A. y = x + 3

B. y = x + 5

C. y = 3x + 7

D. y = 3x + 17

Page 6: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#5 Unit 2This equation can be used to find h, the number of hours it takes Flo and Bryan to mow their lawn.How many hours will it take them to mow their lawn?

A. 6

B. 3

C. 2

D. 1

Least Common Denominator

6

2h + h = 6

13 6

h h

6 6

1 1

13 6

h h

6 6 6

3 6

h h

3h = 6h = 2

3h 63 3

=

Page 7: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#6 Unit 2This equation can be used to determine how many miles apart the two communities are. What is m, the distance between the two communities?

A. 0.5 miles

B. 5 miles

C. 10 miles

D. 15 miles

Least Common Denominator

20

2m = m + 10

5.0515515

mm

0.510 20

m m

20 20

1 1

–m –m

m = 10

0.510 20

m m

20 20 10

10 20

m m

Page 8: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#7 Unit 2For what values of x is the inequality true?

A. x < 1

B. x > 1

C. x < 5

D. x > 5

Least Common Denominator

3

2 + x > 3–2 –2

2 1

3 3

x

2 1

3 3

x

3 3

1 1

6 3 3

3 3

x

x > 1

Page 9: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#8

Unit 2 A manager is comparing the cost of buying ball capswith the company emblem from two different companies.

A. 10 capsB. 20 capsC. 40 capsD. 100 caps

•Company X charges a $50 fee plus $7 per cap.•Company Y charges a $30 fee plus $9 per cap.

For what number of ball caps (b) will themanager’s cost be the same for both companies?

Cost Formula: Company X

CX = 7b + 50Cost Formula: Company Y

CY = 9b + 30CX = CY

7b + 50 = 9b + 30 (Subtract 7b on both sides)

50 = 2b + 30 (Subtract 30 on both sides)

20 = 2b (Divide 2 on both sides)

10 = b

Page 10: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#9Unit 2A shop sells one-pound bags of peanuts for $2 and

three-pound bags of peanuts for $5. If 9 bags are purchased for a total cost of $36, how many three-pound bags were purchased?

Let x = # of one-pound bagsLet y = # of three-pound bags

Method #1Substitution

(Total number of bags)x + y = 9Equation #1:(Total value of bags)2x + 5y = 36Equation #2:

Solve Equation #1 for x

x + y = 9–y –y

x = 9 – y

Substitute x = 9 – y into Equation #2

2(9 – y) + 5y = 3618 – 2y + 5y = 36

18 + 3y = 363y = 18y = 6

6 three-poundbags

Page 11: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#9Unit 2A shop sells one-pound bags of peanuts for $2 and

three-pound bags of peanuts for $5. If 9 bags are purchased for a total cost of $36, how many three-pound bags were purchased?

Let x = # of one-pound bagsLet y = # of three-pound bags

Method #2Elimination

(Total number of bags)x + y = 9Equation #1:(Total value of bags)2x + 5y = 36Equation #2:

Multiply Equation #1 by –2

–2(x + y) = –2(9)

Add New Equation #1 and Equation #2

3y = 18

y = 66 three-pound

bags

–2x – 2y = –18

–2x – 2y = –182x + 5y = 36

(New Equation #1)

Page 12: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#10 Unit 2Which graph represents a system of linear equations that has multiple common coordinate pairs?

A.

C. D.

Has onecommon

coordinatepair

Has onecommon

coordinatepair

Has nocommon

coordinatepairs

Multiplecommon

coordinatepairs

(Two linesoverlap)

B.

Page 13: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#11 Unit 2Which graph represents x > 3 ?

A.

B.

C.

D.

x > 3

x > 3

x < 3

x < 3

Page 14: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#12 Unit 2Which pair of inequalities is shown in the graph?

A. y > –x + 1 and y > x – 5

B. y > x + 1 and y > x – 5

Line 1Line 2

Both given inequalities haveslopes equal to positive one.This is a contradiction to theslope of Line 1 being negative.

NoteLine 1 graph has a negative slope.Line 2 graph has a positive slope.

Page 15: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#12 Unit 2Which pair of inequalities is shown in the graph?

C. y > –x + 1 and y > –x – 5

D. y > x + 1 and y > –x – 5

Line 1Line 2

NoteLine 1 graph has a negative slope.Line 2 graph has a positive slope.

Both given inequalities haveslopes equal to negative one.This is a contradiction to theslope of Line 2 being positive.

Line 2 has a positive slopewith a negative y-intercept.However, the line y > x + 1has a positive slope, but they-intercept is positive.

Page 16: Coordinate Algebra Practice EOCT Answers Unit 2. #1 Unit 2 Which equation shows ax – w = 3 solved for w ? –ax ax – w = 3 –w = 3 – ax –ax = –1 –w 3 – ax

#12 Unit 2Which pair of inequalities is shown in the graph?

A. y > –x + 1 and y > x – 5

B. y > x + 1 and y > x – 5

Line 1Line 2

Both given inequalities haveslopes equal to positive one.This is a contradiction to theslope of Line 1 being negative.

NoteLine 1 graph has a negative slope.Line 2 graph has a positive slope.