Slide 1 - schsalgebra

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7.4 HW Answers
1. (4, -1)
6. (19, 16)
2. (5, 3)
7. (5, 6)
3. (-½, -2)
8. (-7, -12)
4. (9, -3)
9. (2, 1)
5. (-10, -5)
10. (4, 4)
10x + 2y = 155
12x + 3y = 189
$14.50 balls
$5 bags
2
y  x2
3
(-2, -2)
x = -19 + 9
5(y + 9) – 3y = 7
y = -19
x = -10
(-10, -19)
5x  6  7
7  5 x  6  7
1
13
 x
5
5
-1/5
0
13/5
x3 8
x  3  8 or x  3  8
x  5 or x  11
-11
0
5
Scoring Your Homework
• Count how many problems you missed
or didn’t do
•
•
•
•
•
0-1 missed = 10
2-3 missed = 9
4-5 missed = 8
6-7 missed = 7
8-9 missed = 6
• 10-11 missed = 5
• 12-13 missed = 4
• 14-15 missed = 3
• 16-17 missed = 2
• 18-19 missed = 1
• 20-21 missed = 0
Table of Contents
Topic
...
7.1
7.2
7.3
7.4
7.5
Page #
Solve Systems By Graphing
Solve Systems By Substitution
Solve Systems By Combination
Multiplication w/ Combination
System Word Problems
76
77
78
79
80
1. You are selling tickets to a play. The
price of a student ticket is $5, and the
price of an adult ticket is $8. You sell
556 tickets and you collect $3,797. How
many adult and student tickets did you
sell?
Step 1: Choose your variables
S = student ticket
a = adult ticket
1. You are selling tickets to a play. The
price of a student ticket is $5, and the
price of an adult ticket is $8. You sell
556 tickets and you collect $3,797. How
many adult and student tickets did you
sell?
Step 2: Write two equations:
A) One that represents the value of the items
5s + 8a = 3797
B) One that represent the quantity of items
s + a = 556
Step 3: Solve the system using substitution or
elimination
5s + 8a = 3797
(–8) s +
a = 556
217 + a = 556
a = 339
+
5s + 8a = 3797
-8s – 8a = -4448
-3s
= -651
s = 217
2. A bag contains dimes and nickels.
There are 18 coins in the bag. The value
of the coins is $1.25. How many of each
type of coin is in the bag?
Step 1: Choose your variables
d = # dimes
n = # nickels
2. A bag contains dimes and nickels.
There are 18 coins in the bag. The value
of the coins is $1.25. How many of each
type of coin is in the bag?
Step 2: Write two equations:
A) One that represents the value of the items
0.10d + 0.05n = 1.25
B) One that represent the quantity of items
d + n = 18
Step 3: Solve the system using substitution or
elimination
(100)
0.10d + 0.05n = 1.25
(–5)
d + n = 18
7 + n = 18
n = 11
+
10d + 5n = 125
-5d – 5n = -90
5d
= 35
d=7
3. Kylie has 4.50 in dimes and quarters. She has
3 more dimes than quarters. How many
quarters does she have?
d = # dimes
q = # quarters
0.10d + 0.25q = 4.50
d = q + 3 = 12 + 3 = 15
0.10(q + 3) + 0.25q = 4.50
0.10q + 0.30 + 0.25q = 4.50
0.35q + 0.30 = 4.50
0.35q = 4.20
q = 12
4. An adult ticket to a school play costs $5 and a
student ticket costs $3. A total of $460 was
collected from the sale of 120 tickets. How
many student tickets were purchased?
a = adult ticket
s = student ticket
5a + 3s = 460
a + s = 120
5a + 3s = 460
(–3) a + s = 120
+
5a + 3s = 460
-3a – 3s = -360
2a
50 + s = 120
s = 70
= 100
a = 50
5. A drummer goes to Guitar Center and buys
drum sticks and brushes. The wood sticks that
he buys are $10.50 a pair, and the brushes are
$24 a pair. He buys twice as many pairs of
wood sticks as brushes, and ends up spending
a total of $90. How many pair of sticks and
brushes did he buy?
d = drum sticks
b = brushes
10.50d + 24b = 90
d = 2b
10.50d + 24b = 90
d = 2b = 2(2) = 4
10.50(2b) + 24b = 90
21b + 24b = 90
45b = 90
b=2
6. The sophomore class is selling pretzels and
popcorn at a school event to raise money for a
dance. They charge $2.50 for a bag of popcorn
and $2 for a pretzel. They collect $336 during
the event. They sell twice as many bags of
popcorn as pretzels. How many pretzels do
they sell?
z = pretzels
p = popcorn
2.50p + 2z = 336
p = 2z
2.50p + 2z = 336
p = 2z = 2(48) = 96
2.50(2z) + 2z = 336
5z + 2z = 336
7z = 336
z = 48
7. You went to Home Depot and bought two types
of plants. One type was a flowering plant while
the other was non-flowering. The flowering
plant cost $3.20 each and the non-flowering
plant cost $1.50 each. You purchased a total of
24 plants for $49.60. How many of each type of
plant did he buy?
f = flower plant
n = non-flower
3.20f + 1.5n = 49.60
f + n = 24
(10)
3.20f + 1.5n = 49.60
(–15)
f + n = 24
8 + n = 24
n = 16
32f + 15n = 496
+
-15f – 15n = -360
17f
= 136
f=8

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