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CHAPTER 2
Applications of Linear Models
System of Linear equations with Two
Variables
Example
x+y=2
x–y=3
By substitution or elimination you will find
x and y values.
Ch 2
Systems of Linear Equations (Pg 86)
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Average weight of a thrush : t
Average weight of a robin : r
(Weight of thrushes) + ( weight of robins) = total weight
Thus 3t + 6r = 48
5t + 2r = 32
This pair of equations is an example of a linear system of
two equations in two unknowns ( 2 x 2 linear system)
• A solution to the system is an ordered pair of numbers (t, r)
that satisfies both equations in the system
20
5t +2r = 32
10
5 (4) + 2 (6) = 32, True
(4, 6)
3t + 6r = 48
5
3(4) + 6(6) = 48, True
Conclusion : Both equations are true, so average weight of a thrush is 4 ounces,
and the average weight of a robin is 6 ounces
Using Graphing Calculator (Pg 88)
Enter Y1 = (21.06 – 3x)/ -2.8
Y2 = (5.3 – 2x)/1.2
Enter
press zoom 6
Enter
Final Graph
then press 2nd , calc
Enter
Inconsistent and Dependent Systems
Dependent system
Inconsistent system
Consistent and Independent system
( Infinitely many solutions) ( parallel lines and no solution) Intersect in one point and
Exactly one solution
Pg 90 Example 2
Press Y1 = -X + 5
Y2 = -X + 1.5
Enter Graph
10
9.4
-9.4
Example 4
Enter equation
- 10
Press window and enter
Press graph and calc
• Step 1
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• Step 2
Example 4, Pg 91
Fraction of a cup of oats needed:
Fraction of a cup of wheat needed:
x
y
Cups
Grams of
Protein
Per Cup
Grams of
protein
Oats
x
11
11x
Wheat
y
8.5
8.5y
Mixture
1
-
10
First equation x + y = 1
Second Equation 11x + 8.5y = 10
Solve the system of graphing using Graphing calculator
y=-x+1
y = (10 – 11x) /8.5
X min = 0 Xmax = 0.94
Ymin = 0 Ymax = 1
Francine needs 0.6 cups of oats and 0.4 cups of wheat
Ex 2.1, Pg 97, No. 19
a)
b)
c)
d)
f)
Supply equation y= 50x
Demand equation y = 2100 – 20x
The graph of y = 2100 – 20x has y intercept (0, 2100) and xintercept (105, 0)
Xmin = 0, Xmax = 120
Ymin = 0, Ymax = 2500
Press Y enter equations
a)
b)
Press window, enter values ,
press 2nd and table,
press graph and trace
The equilibrium price occurs at the intersection point (30,
1500) in the above graph
To verify
Y = 50(30) = 1500
Y = 2100 – 20(30) = 1500
Yasuo should sell the wheat at 30 cents per bushel
and produce 1500 bushels
2.2 Solutions of Systems by Algebraic Methods
By Substitution
Example 1 (pg 99)
Step 1 Number of standard sleeping bags: x
Number of down-filled sleeping bags:
y
Step 2 Staci needs twice as many standard
model as down-filled x = 2y
Also, the total number of sleeping bags is 60
x + y = 60
Step 3 Substitute x = 2y in second equation
2y + y = 60, 3y = 60, y = 20
Solving for y we find y = 20, x= 2(20) = 40
The solution to the system is x = 40, y = 20
Staci should order 40 standard sleeping bags and 20 down-filled
bags
Solutions of Systems by Algebraic Methods
By Elimination Ex 3, pg 101
2x + 3y = 8
3x – 4y = -5
Multiply first equation by 3 and second equation by –2
6x + 9y = 24
-6x + 8y = 10
Add
17 y = 34
y=2
Substitute in first equation 2x + 6 = 8
2x = 8-6
2x = 2
x=1
The ordered pair (1, 2)
Solve by Linear Combination( Ex – 2.2, No 12, pg 106 )
• 2p + 8q = 4
3
9
3
• P = 2+q
3
2
6p + 8q = 12
( Multiply the first equation by 9)
• 2p = 12 + 3q
(the second equation by 6)
Standard form
• 6p + 8q = 12
2p – 3q = 12
• 6p + 8q = 12
-6p + 9Q = -36 (Multiply the second equation by –3)
• Add the equations
• 17q = -24, q = -24/17
• Substitute q
• 2p = 12 + 3q
• 2p = 12 + 3( -24/17)
• 2p = 12 – 72/17
• 2p = 132/17, P = 66/17 The solution p = 66/17, q = -24/17
Ex 2.2, Pg 106, No 21
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Let s represent the salinity in percent
Let M represent the temperature of maximum density
From the ordered pairs (0, 4) and (15, 0.8)
The M-intercept is 4 and the slope is
m = 0.8 – 4 = -3.2 = - 16
15 – 0
15
75
Hence M = -16/75 s + 4
b) Let s represent the salinity in percent and let F represent
the freezing point. Form the ordered pairs (0, 0) and (15, 0.8). The F-intercept is 0 and the slope is m = -0.8/15 = 4/75
• Hence F = -4/75 s
• Graph
• d) Let F = M and then solve for s
• -16s + 300 = -4s( multiply by 75)
• 300 = 12s
• S = 25
Substitute the value in one of the equation M = - 16/75 (25) + 4
= -1 1/3 The salinity is 25% and the freezing point is – 1 1/3
C
2.5 Linear Inequalities in Two Variables
8000
y > - x + 10000
x + y > 10000
4000
x + y = 10000
4000
8000
12000
Use a test point
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3x – 2y < 6
First graph the line 3x – 2y = 6
The intercepts are (2, 0) and (0, -3)
Next choose a test point. Since (0,0) does not
lie on the line, we choose it as test point
• 3(0) – 2(0) < 6 True So we shade the half
plane that contains the test point
3x – 2y < 6
3x – 2y = 6
Ex 2.5, No 33, Pg 139Graph each system of
inequalities and find the coordinates of the vertices
x+y>3
2y < x + 8
2y + 3x < 24
x > 0, y > 0
• First graph x + y = 3
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2y = x + 8
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2y + 3x = 24 ( y= 0, X = 8; Finally substitute x = 8, y = 4 )
y
(4, 6)
5
(0, 4)
(0, 3)
(3, 0)
(8, 0) 10
x
Ex 2.5, No 37( Pg 139)
Let x represent the number of student tickets sold
y represent the number of faculty tickets sold
The information that student tickets cost $1, faculty
tickets cost $2, and the receipts must be atleast
$250, can be stated in the inequality x + 2y > 250
So positive no of Tickets are sold
x> 0 and y> 0
The system of inequalities is x + 2y > 250
x + 2y = 250
x > 0, y > 0 150
2y = 250 – x
y = 125 – x/2
100
50
y = 125 – x/2
0
4
8
12