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 t +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 2 nd, 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 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 CupsGrams of Protein Per Cup Grams of protein Oatsx1111x Wheaty8.58.5y Mixture1 -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) Supply equation y= 50x b) Demand equation y = 2100 – 20x c) The graph of y = 2100 – 20x has y intercept (0, 2100) and x- intercept (105, 0) d) Xmin = 0, Xmax = 120 f) Ymin = 0, Ymax = 2500 a) The equilibrium price occurs at the intersection point (30, 1500) in the above graph b) 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 Press Y enter equations Press window, enter values, press 2 nd and table, press graph and trace

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 = P = 2 + q 3 2 6p + 8q = 12 ( Multiply the first equation by 9) 2p = q (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 = q 2p = ( -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 = – 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 = -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 x + y > x + y = y > - x

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 > 3 2y 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 > x + 2y = 250 2y = 250 – x y = 125 – x/