1. Coordinate Algebra
Practice
EOCT Answers
Unit 2

2. = ax – w = 3 –w = 3 – ax –w 3 – ax –1 w = –3 + ax w = ax – 3 –ax –ax#1
Unit 2
Which equation shows ax – w = 3
solved for w ?
ax – w = 3
–ax
–ax
A. w = ax – 3
B. w = ax + 3
C. w = 3 – ax
D. w = 3 + ax
–w = 3 – ax = –1
–w – ax
w = –3 + ax
w = ax – 3

3. Least Common Denominator#2
Unit 2
Which equation is equivalent to ?
Least Common Denominator 8
A. 17x = 88
B. 11x = 88
C. 4x = 44
D. 2x = 44
14x – 3x = 88
11x = 88

4. Method #1 Method #2 4n = 2(t – 3) 4n = 2(t – 3) 4n = 2t – 6 2#3
Unit 2
Which equation shows 4n = 2(t – 3)
solved for t ?
Method #1
Method #2
4n = 2(t – 3)
4n = 2(t – 3)
4n = 2t – 6 2
4n 2(t – 3) = +6 +6
4n + 6 = 2t
2n = t – 3 +3 +3
4n t 2 =
2n + 3 = t
2n + 3 = t

5. 6(x + 4) = 2(y + 5) 6x + 24 = 2y + 10 –10 –10 6x + 14 = 2y 6x + 14 2y#4
Unit 2
Which equation shows 6(x + 4) = 2(y + 5)
solved for y ?
6(x + 4) = 2(y + 5)
A. y = x + 3
B. y = x + 5
C. y = 3x + 7
D. y = 3x + 17
6x = 2y + 10
–10
–10
6x = 2y
6x y 2 =
3x + 7 = y

6. Least Common Denominator#5
Unit 2
This 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?
Least Common Denominator 6
A. 6
B. 3
C. 2
D. 1 3h 3 =
2h + h = 6
h = 2
3h = 6

7. Least Common Denominator#6
Unit 2
This 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 –m –m
m =

8. Least Common Denominator#7
Unit 2
For what values of x is the inequality true?
Least Common Denominator 3
A. x < 1
B. x > 1
C. x < 5
D. x > 5
2 + x > 3 –2 –2
x > 1

9. CX = 7b + 50 CY = 9b + 30 CX = CY 7b + 50 = 9b + 30 50 = 2b + 30
Unit 2
A manager is comparing the cost of buying ball caps
with the company emblem from two different companies. #8
Company X charges a $50 fee plus $7 per cap.
Company Y charges a $30 fee plus $9 per cap.
A. 10 caps
B. 20 caps
C. 40 caps
D caps
For what number of ball caps (b) will the
manager’s cost be the same for both companies?
Cost Formula: Company X
CX = 7b + 50
Cost Formula: Company Y
CY = 9b + 30
CX = 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

10. Method #1 Substitution x + y = 9 2x + 5y = 36 2(9 – y) + 5y = 36
A 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?
Unit 2 #9
Method #1
Substitution
Let x = # of one-pound bags
Let y = # of three-pound bags
(Total number of bags)
x + y = 9
Equation #1:
(Total value of bags)
2x + 5y = 36
Equation #2:
Substitute x = 9 – y into Equation #2
Solve Equation #1 for x
x + y = 9
2(9 – y) + 5y = 36
18 – 2y + 5y = 36 –y
18 + 3y = 36
x = 9 – y
3y = 18
6 three-pound
bags
y = 6

11. Method #2 Elimination x + y = 9 2x + 5y = 36 –2x – 2y = –18
A 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?
Unit 2 #9
Method #2
Elimination
Let x = # of one-pound bags
Let y = # of three-pound bags
(Total number of bags)
x + y = 9
Equation #1:
(Total value of bags)
2x + 5y = 36
Equation #2:
Add New Equation #1 and Equation #2
Multiply Equation #1 by –2
–2x – 2y = –18
2x + 5y = 36
–2(x + y) = –2(9)
–2x – 2y = –18
3y = 18
(New Equation #1)
6 three-pound
bags
y = 6

12. #10
Unit 2
Which graph represents a system of linear equations that has multiple common coordinate pairs? A. B.
Multiple
common
coordinate
pairs
(Two lines
overlap) 
Has one
common
coordinate
pair D. C.
Has no
common
coordinate
pairs
Has one
common
coordinate
pair 

13. x > 3 x < 3 x > 3 x < 3 Which graph represents x > 3 ?
#11
Unit 2
Which graph represents x > 3 ? A.
x > 3
x < 3 B. C.
x > 3 D.
x < 3

14. Which pair of inequalities is shown in the graph?
#12
Unit 2
A. y > –x + 1 and y > x – 5
B. y > x + 1 and y > x – 5
Line 1
Line 2
Both given inequalities have
slopes equal to positive one.
This is a contradiction to the
slope of Line 1 being negative.
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.

15. Which pair of inequalities is shown in the graph?
#12
Unit 2
C. y > –x + 1 and y > –x – 5
D. y > x + 1 and y > –x – 5
Both given inequalities have
slopes equal to negative one.
This is a contradiction to the
slope of Line 2 being positive.
Line 1
Line 2
Line 2 has a positive slope
with a negative y-intercept.
However, the line y > x + 1
has a positive slope, but the
y-intercept is positive.
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.

16. Which pair of inequalities is shown in the graph?
#12
Unit 2
A. y > –x + 1 and y > x – 5
B. y > x + 1 and y > x – 5
Line 1
Line 2
Both given inequalities have
slopes equal to positive one.
This is a contradiction to the
slope of Line 1 being negative.
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.