Transcript Section 8.4 Chapter 8 Systems of Linear Equations in Two Variables

Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Exercise #13
Set up a system of linear equations in two
variables, and solve for the unknown
quantities.
Shanelle invested $10,000 and at the end of
1 year she received $805 in interest. She
invested part of the money in an account
earning 10% simple interest and the
remaining money in an account earning
7% simple interest. How much did she
invest in each account?
Principal
invested
Interest earned
10%
Account
x
.10x
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
x + y = 10,000
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
x + y = 10,000
.10x + .07y = 805
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
x + y = 10,000
100 .10x + .07y = 805


7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
x + y = 10,000
10x + 7y = 80,500
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned

10%
Account
7%
Account
x
.10x
– 7 x + y = 10,000
10x + 7y = 80,500

Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
– 7x – 7y = – 70,000
10x + 7y = 80,500
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
– 7x – 7y = – 70,000
10x + 7y = 80,500
3x = 10,500
x = 3,500
x + y = 10,000
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
.10x
– 7x – 7y = – 70,000
10x + 7y = 80,500
3x = 10,500
x = 3,500
3500 + y = 10,000
y = 6500
7%
Account
Total
y
10,000
.07y
805
Principal
invested
Interest earned
10%
Account
x
7%
Account
.10x
Shanelle invested $3500
in the 10% account and
$6500 in the 7% account.
Total
y
10,000
.07y
805
Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Exercise #19
Set up a system of linear equations in two
variables, and solve for the unknown
quantities.
How much 50% disinfectant solution
must be mixed with a 40% disinfectant
solution to produce 25 gal of a 46%
disinfectant solution?
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
x + y = 25
40%
Mixture
46%
Mixture
.40y
.46 25
y
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
y
x + y = 25
.50x + .40y = .46 25
 
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
x + y = 25
100 .50x + .40y = .46 25

y
 
50x + 40y = 46 25 
50x + 40y = 1150
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
x + y = 25
50x + 40y = 1150
40%
Mixture
46%
Mixture
.40y
.46 25
y
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
–50 x + y = 25 
50x + 40y = 1150
40%
Mixture
46%
Mixture
.40y
.46 25
y
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
– 50x – 50y = – 1250
50x + 40y = 1150
y
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
y
– 50x – 50y = – 1250
50x + 40y = 1150
– 10y = – 100
y = 10
x + y = 25
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
y
– 50x – 50y = – 1250
50x + 40y = 1150
– 10y = – 100
y = 10
x + 10 = 25
x = 15
25
 
50%
Amount of Mixture
x
solution
Amount of
.50x
disinfectant
40%
Mixture
46%
Mixture
.40y
.46 25
y
15 gallons of the 50% mixture
should be mixed with 10 gallons
of the 40% mixture.
25
 
Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Exercise #23
Set up a system of linear equations in two
variables, and solve for the unknown
quantities.
A jar of face cream contains 18%
moisturizer, and another is 24%
moisturizer. How many ounces of
each should be combined to get
12 oz of a cream that is 22%
moisturizer?
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
x + y = 12
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
x + y = 12
.18x + .24y = .22 12
 
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
x + y = 12
100 .18x + .24y = .22 12

 
18x + 24y = 22 12 
18x + 24y = 264
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
x + y = 12
18x + 24y = 264
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer

18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y

– 18 x + y = 12
18x + 24y = 264
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
– 18x – 18y = – 216
18x + 24y = 264
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
– 18x – 18y = – 216
18x + 24y = 264
6y = 48
y =8
x + y = 12
22%
Moisturizer
12
 
.22 12
Amount of
face cream
Amount of
moisturizer
18% Moisturizer 24%
Moisturizer
x
y
.18x
.24y
– 18x – 18y = – 216
18x + 24y = 264
6y = 48
y =8
x + 8 = 12
4 oz of 18% moisturizer;
x =4
8 oz of 24% moisturizer
22%
Moisturizer
12
 
.22 12
Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Exercise #25
Set up a system of linear equations in two
variables, and solve for the unknown
quantities.
It takes a boat 2 hr to go 16 miles
downstream with the current, and
4 hr to return against the current.
Find the speed of the boat in still
water and the speed of the current.
Distance
Rate
Time
Downstream
16
x +y
2
Return
16
x–y
4
x = speed of boat in still water
y = speed of current
16 = 2 x + y


Distance
Rate
Time
Downstream
16
x +y
2
Return
16
x–y
4
x = speed of boat in still water
y = speed of current
16 = 2 x + y
2x + 2y = 16
2 2x + 2y = 16




4x + 4y = 32

16 = 4 x – y
4x – 4y = 16

Distance
Rate
Time
Downstream
16
x +y
2
Return
16
x–y
4
4x + 4y = 32
4x – 4y = 16
Distance
Rate
Time
Downstream
16
x +y
2
Return
16
x–y
4
4x + 4y = 32
4x – 4y = 16
8x = 48
x =6
2x + 2y = 16
Distance
Rate
Time
Downstream
16
x +y
2
Return
16
x–y
4
4x + 4y = 32
4x – 4y = 16
8x = 48
x =6

2 6 + 2y = 16
12 + 2y = 16
2y = 4
y =2
The speed of the boat in still
water is 6 mph, and the
speed of the current is 2
Chapter 8
Systems of Linear Equations in Two Variables
Section 8.4
Exercise #31
Set up a system of linear equations in two
variables, and solve for the unknown
quantities.
Debi has $2.80 in a collection of
dimes and nickels. The number
of nickels is five more than the
number of dimes. Find the number
of each type of coin.
Number of coins
Value of coins
Dimes
Nickels
d
n
.10d
.05n
n =5+d
Total
2.80
Number of coins
Value of coins
Dimes
Nickels
d
n
.10d
.05n
n =5+d
.10d + .05n = 2.80
Total
2.80
Number of coins
Value of coins

Dimes
Nickels
d
n
.10d
.05n
n =5+d
100 .10d + .05n = 2.80
10d + 5n = 280

Total
2.80
Number of coins
Value of coins

Dimes
Nickels
d
n
.10d
.05n
n =5+d
100 .10d + .05n = 2.80


10d + 5 5 + d = 280
10d + 25 + 5d = 280
15d + 25 = 280
15d = 255
d = 17

Total
2.80
Number of coins
Value of coins

Dimes
Nickels
d
n
.10d
.05n
Total
2.80
n =5+d
100 .10d + .05n = 2.80



10d + 5 5 + d = 280
10d + 25 + 5d = 280
15d + 25 = 280
15d = 255
d = 17
n =5+d
Number of coins
Value of coins

Dimes
Nickels
d
n
.10d
.05n
Total
2.80
n =5+d
100 .10d + .05n = 2.80



10d + 5 5 + d = 280
10d + 25 + 5d = 280
15d + 25 = 280
15d = 255
d = 17
n = 5 + 17
n = 22
Number of coins
Value of coins

Dimes
Nickels
d
n
.10d
.05n
Total
2.80
n =5+d
100 .10d + .05n = 2.80



10d + 5 5 + d = 280
10d + 25 + 5d = 280
15d + 25 = 280
15d = 255 n = 5 + 17
Debi has 17 dimes and 22 nickels.