Download Sections 3

Sections 3.1 and 3.2---Solving Systems of Equations
Part 1: General Information
(1) A system of equations is two or more equations with the
same variable.
Example: A: 2x – 3y = 6
B: 3y = 2x + 4
(2) The solution for a system of equations is the point
(ordered pair) where the graphs intersect—if they do
intersect.
(3) If the graphs do not intersect then there are no solutions.
(4) If the graphs coincide, then there are infinitely many
solutions.
Part 2: Complete the table. Show initial steps in the table.
Section 3-1 and 3-2 Notes
page 2
Part 3: Applications. Use a system of equations to solve. Write “lets” and
equations, then solve by any method.
1. A service club is selling copies of their holiday cookbook to raise funds
for a project. The printer’s set-up charge is $200, and each book costs
$2 to print. The cookbooks will sell for $6 each. How many cookbooks
must the members sell before they make a profit? (What is the break
even point?)
Let x = the number of cookbooks they must sell
C(x) = cost in dollars
R(x) = revenue in dollars
C(x) = 2x + 200
R(x) = 6x
y2 = 6x
400
350
y1= 2x + 200
300
dollars
250
200
150
100
50
-10
10 20 30 40 50 60 70 80 90 100
number of cookbooks
2. From two kinds of corn meal, one costing $0.65/kg and the other
$0.85/kg, a farmer wishes to make a 100 kg mixture costing $0.76/kg.
How many kg of each kind should he use?
Kind of Cornmeal Number of Kg Cost
0.65x
$0.65
x
0.85y
$0.85
y
$0.76
100
0.76(100)
x + y = 100
0.65x + 0.85y = 76
x = 45 kg of the $0.65 cornmeal and y = 55 kg of
the $0.85 cornmeal
Section 3-1 and 3-2 Notes
page 3
3. Joni invested $16,000 in two different accounts which pay 5% and 7 ½%
annual interest, respectively. After one year she had earned $1094.13
in interest. How much did she invest at each rate?
Kind of
Account
5%
7 ½%
Amount Invested in this Interest
Account ($)
Earned ($)
x
0.05x
y
0.075y
16000
1094.13
x + y = 16000
0.05x + 0.075y = 1084.13
Joni invested $4234.80 at 5% and $11,765.20 at
7.5%.
4. James has 250 coins in his piggy bank. If the bank contains only nickels
and dimes and the total value of the coins is $16.45, how many of each
type coin does he have?
Let n = # of nickels (0.05 each)
A table would work
nicely here!!
d = # of dimes (0.10 each)
n + d = 250
0.05n + 0.10d = 16.45
James has 171 nickels and 79 dimes.
5. The sum of a smaller number and twice a larger number is 189. If one
of the numbers exceeds the other number by 72, find the two numbers.
Let x = the smaller number
y = the larger number
x + 2y = 189
x – y = 72
The smaller number is 15 and the larger number is 87.
6. Salt solutions of 60% and 15% are mixed to produce a 450 cm3 mixture
of a 25% solution. How much of each solution is required?
“Kind” of
Number of cm3 of
Number of cm3 of
Solution
solution
salt
60%
x
0.60x
15%
y
0.15y
25%
450
.25(450)
x + y = 450
.60x + .15y = 112.5
One hundred cm3 of the 60% solution should be mixed with 350 cm3 of
the 15% solution.
Section 3-1 and 3-2 Notes and HW
7.
Let t = the ten’s digit and
digit
t + u = 14
and
u = ½ t + 2;
6
The number is 86.
8.
“Kind” of
Number of lbs. of
Solution
copper alloy
page 4
u = the one’s
t = 8 and u =
Number of lbs.
of copper
12%
x
0.12x
32%
y
0.32y
20%
80
.2 (80) = 16
x + y = 80
and 0.12x + 0.32y = 16; x = 48 and
y = 32.
Forty-eight pounds of 12% solution and 32 pounds
of 32% solution should be used.
9. Let t = the ten’s digit and u = the one’s
digit
t+u=9
and
10u + t = 10t + u – 9;
t = 5 and u = 4
The number is 54.
Part 4:
1. Let x = the larger number and y = the smaller number
x + y = 62 and x = y + 8; x = 35 and y = 27
2. Let t = the ten’s digit and u = the unit’s digit.
t + u = 10; 5t – 3u = 18; t = 6 and u = 4.
The number is 65=4.
3. Let L = the length and w = the width; 2L + 2w = 56 and
w + 1/2 L = 18 The length is 20 feet and the width is 8 feet.
4. Let A = the cost of an apple and R = the cost of an orange; 9A +7R =
7.30 and 4A + 11R = 6.40. An apple costs $0.50 and an orange costs
$0.40.
5.
Kind of Account Amount Invested in this Account ($) Interest Earned ($)
4.5%= 0.045
x
0.045x
3% = 0.03
y
0.03y
15600
588
x + y = 15600
0.045x + 0.03y = 588
Andy invested $8000.00 at 4.5% and $7600.00 at 3%.
6. Let x = the larger integer and y = the smaller integer
x + y = 188 and x – y = 34. The integers are 111 and 77.
Section 3-1 and 3-2 Notes and HW
page 5
7. Let t = ten’s digit and u = the one’s digit; t + u =13 and
u = 2t – 2; u = 8 and t = 5. The number is 58.
8. Let L = the length and w = the width; 2L + 2w = 44 and
L + 2 w = 31. The length is 13 in. and the width is 9 in.
9.
Kind of Ticket Number of Tickets Cost
$3.00 (children)
x
3x
$4.50 (all else)
y
4.50y
Total
225
937.50
x + y = 225 and 3x + 4.5y = 937.5 There were 50 children’s tickets sold and
175 other tickets sold.
10.
Kind of Grain Number of lbs. Cost
$0.65
x
0.65x
$0.90
y
0.9y
$0.80
20
0.8(20)
x + y = 20 and 0.65x + 0.9y = 16. Add 8 lbs of the $0.65 grain and 12 lbs.
of the $0.90 grain.
11. Let x = one number and y = the other number
x = 3y +8 and x + y =68. The numbers are 53 and 15.
12. Let t = ten’s digit and u = the one’s digit; t + u =10 and
5t +6u = 57 ; u = 7 and t = 3. The number is 37.
13. Let x = # of $2.25 tickets and y = # of $3.50 tickets
x + y = 165 and 2.25x + 3.5y = 482.5; There were 76 $2.25 tickets
sold and 89 $3.50 tickets sold.
14. Let x = the # kg of $1.25 candy and y = the # kg of $1.65 candy
x + y = 10 and 1.25x + 1.65y = 14. There was 6.25 kg of the $1.25
candy sold and 3.75 kg of the $1.65 candy sold.
15. Let x = # of pairs of $20 shoes sold and y = # pairs of $25 shoe
sold. X + y = 55 and 20x + 25y = 1250 There were 25 pairs of the
$20 shoes sold and 30 pairs of the $25 shoes sold.
16. Let x = # of student tickets sold and y = # of adult tickets sold
x + y = 323 and 1.25x + 1.75y = 478.75. There were 173 student
tickets sold and 150 adult tickets sold.