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)• 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

5

20

10

5t +2r = 32 5 (4) + 2 (6) = 32, True

3t + 6r = 48 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

(4, 6)

Using Graphing Calculator (Pg 88)

Enter Y1 = (21.06 – 3x)/ -2.8 Y2 = (5.3 – 2x)/1.2 press zoom 6 then press 2nd , calc

Enter Enter Enter

Final Graph

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

Press Y1 = -X + 5 Enter Graph Y2 = -X + 1.5

- 10

10

9.4-9.4

Pg 90 Example 2

Enter equation Press window and enter Press graph and calc

Example 4

Example 4, Pg 91• Step 1 Fraction of a cup of oats needed: x• Fraction of a cup of wheat needed: y• Step 2

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 = 1Second Equation 11x + 8.5y = 10Solve the system of graphing using Graphing calculatory = - x + 1y = (10 – 11x) /8.5

X min = 0 Xmax = 0.94Ymin = 0 Ymax = 1

Francine needs 0.6 cups of oats and 0.4 cups of wheat

Ex 2.1, Pg 97, No. 19a) Supply equation y= 50xb) Demand equation y = 2100 – 20xc) The graph of y = 2100 – 20x has y intercept (0, 2100) and x-intercept (105, 0)d) Xmin = 0, Xmax = 120f) Ymin = 0, Ymax = 2500

a) The equilibrium price occurs at the intersection point (30, 1500) in the above graphb) To verify Y = 50(30) = 1500 Y = 2100 – 20(30) = 1500

Yasuo should sell the wheat at 30 cents per busheland produce 1500 bushels

Press Y enter equations Press window, enter values , press 2nd and table, press graph and trace

2.2 Solutions of Systems by Algebraic MethodsBy 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 MethodsBy Elimination Ex 3, pg 101

2x + 3y = 83x – 4y = -5Multiply first equation by 3 and second equation by –26x + 9y = 24-6x + 8y = 10Add17 y = 34y = 2Substitute 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• 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 = 25Substitute 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

4000 8000 12000

8000

4000

x + y > 10000

x + y = 10000

y > - x + 10000

Use a test point

• 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 > 32y < x + 8

2y + 3x < 24x > 0, y > 0• First graph x + y = 3

• 2y = x + 8• 2y + 3x = 24 ( y= 0, X = 8; Finally substitute x = 8, y = 4 )

(3, 0) (8, 0) 10 x

5(0, 4)(0, 3)

(4, 6)y

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 > 0, y > 0 150

100

50

x + 2y = 2502y = 250 – xy = 125 – x/2

y = 125 – x/2

0 4 8 12