coordinate algebra practice eoct answers unit 2...a. y > –x + 1 and y > x – 5 b. y > x...
TRANSCRIPT
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
w = –3 + ax
w = ax – 3
A. w = ax – 3
B. w = ax + 3
C. w = 3 – ax
D. w = 3 + ax
#2 Unit 2 Which 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 3
114 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
#3 Unit 2 Which equation shows 4n = 2(t – 3)
solved for t ?
4n = 2(t – 3)
2n = t – 3
2 4n 2(t – 3)
2 =
+3 +3
2n + 3 = t
4n = 2(t – 3)
4n = 2t – 6 +6 +6
4n + 6 = 2t
2n + 3 = t
Method #1 Method #2
4n + 6 2t 2 2
=
#4 Unit 2 Which equation shows 6(x + 4) = 2(y + 5)
solved for y ?
6(x + 4) = 2(y + 5)
–10
6x + 24 = 2y + 10
–10
6x + 14 = 2y
3x + 7 = y
6x + 14 2y 2 2
=
A. y = x + 3
B. y = x + 5
C. y = 3x + 7
D. y = 3x + 17
#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?
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 = 6 h = 2
3h 6 3 3
=
#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
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
#7 Unit 2 For 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
13 3
x
3 3
1 1
6 3 3
3 3
x
x > 1
#8
Unit 2 A manager is comparing the cost of buying ball caps
with the company emblem from two different companies.
A. 10 caps
B. 20 caps
C. 40 caps
D. 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 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
#9 Unit 2 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?
Let x = # of one-pound bags
Let y = # of three-pound bags
Method #1
Substitution
(Total number of bags) x + y = 9 Equation #1:
(Total value of bags) 2x + 5y = 36 Equation #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 = 36
18 – 2y + 5y = 36
18 + 3y = 36
3y = 18
y = 6 6 three-pound
bags
#9 Unit 2 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?
Let x = # of one-pound bags
Let y = # of three-pound bags
Method #2
Elimination
(Total number of bags) x + y = 9 Equation #1:
(Total value of bags) 2x + 5y = 36 Equation #2:
Multiply Equation #1 by –2
–2(x + y) = –2(9)
Add New Equation #1 and Equation #2
3y = 18
y = 6 6 three-pound
bags
–2x – 2y = –18
–2x – 2y = –18
2x + 5y = 36
(New Equation #1)
#10 Unit 2 Which graph represents a system of linear equations
that has multiple common coordinate pairs?
A.
C. D.
Has one
common
coordinate
pair
Has one
common
coordinate
pair
Has no
common
coordinate
pairs
Multiple
common
coordinate
pairs
(Two lines
overlap)
B.
#11 Unit 2 Which graph represents x > 3 ?
A.
B.
C.
D.
x > 3
x > 3
x < 3
x < 3
#12 Unit 2 Which 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 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.
#12 Unit 2 Which 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 1 Line 2
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.
Both given inequalities have
slopes equal to negative one.
This is a contradiction to the
slope of Line 2 being positive.
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.
#12 Unit 2 Which 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 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.