# 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations CHAPTER 9.1Solving Systems of Linear Equations Graphically.

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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations CHAPTER 9.1Solving Systems of Linear Equations Graphically 9.2Solving Systems of Linear Equations by Substitution 9.3Solving Systems of Linear Equations by Elimination 9.4Solving Systems of Linear Equations in Three Variables 9.5Solving Systems of Linear Equations Using Matrices 9.6Solving Systems of Linear Equations Using Cramer’s Rule 9.7Solving Systems of Linear Inequalities 9

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations Graphically 9.1 1.Determine if an ordered pair is a solution for a system of equations. 2.Solve a system of linear equations graphically. 3.Classify systems of linear equations in two unknowns.

Slide 9- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Determine if an ordered pair is a solution for a system of equations.

Slide 9- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley System of equations: A group of two or more equations. Solution for a system of equations: An ordered set of numbers that makes all equations in a system true.

Slide 9- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Checking a Solution to a System of Equations To verify or check a solution to a system of equations: 1. Replace each variable in each equation with its corresponding value. 2. Verify that each equation is true.

Slide 9- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether each ordered pair is a solution to the system of equations. a. (  3, 2)b) (3, 4) Solution a. (  3, 2) x + y = 7y = 3x – 2  3 + 2 = 7 2 = 3(  3) – 2  1 = 72 =  11 FalseFalse

Slide 9- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued b. (3, 4) x + y = 7y = 3x – 2 3 + 4 = 7 4 = 3(2) – 2 7 = 74 = 4True Because (  3, 2) does not satisfy both equations, it is not a solution for the system. Because (3, 4) satisfies both equations, it is a solution to the system of equations.

Slide 9- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve a system of linear equations graphically.

Slide 9- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A system of two linear equations in two variables can have one solution, no solution, or an infinite number of solutions. The graphs intersect at a single point. The equations have the same slope, the graphs are parallel. The graphs are identical. There are an infinite number of solutions.

Slide 9- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Equations Graphically To solve a system of linear equations graphically: 1. Graph each equation. a. If the lines intersect at a single point, then the coordinates of that point form the solution. b. If the lines are parallel, then there is no solution. c. If the lines are identical, then there are an infinite number of solutions, which are the coordinates of all the points on that line. 2. Check your solution.

Slide 9- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations graphically. Solution Graph each equation. The lines intersect at a single point, (  2, 4). We can check the point in each equation to verify and will leave that to you. y = 2 – x 2x + 4y = 12 (  2, 4)

Slide 9- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations graphically. Solution Graph each equation. The lines appear to be parallel, which we can verify by comparing the slopes. The slopes are the same, so the lines are parallel.

Slide 9- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Classify systems of linear equations in two unknowns.

Slide 9- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Consistent system of equations: A system of equations that has at least one solution. Inconsistent system of equations: A system of equations that has no solution. Dependent linear equations in two unknowns: Equations with identical graphs. Independent linear equations in two unknowns: Equations that have different graphs.

Slide 9- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Classifying Systems of Equations To classify a system of two linear equations in two unknowns, write the equations in slope-intercept form and compare the slopes and y-intercepts. Consistent system: The system has a single solution at the point of intersection. Independent equations: The graphs are different and intersect at one point. They have different slopes. Consistent system: The system has an infinite number of solutions. Dependent equations: The graphs are identical. They have the same slope and same y-intercept. Inconsistent system: The system has no solution. Independent equations: The graphs are different and are parallel. Though they have the same slopes, they have different y- intercepts.

Slide 9- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Returning to a previous example (1) is the system consistent or inconsistent, are the equations dependent or independent, and how many solutions does the system have? The graphs intersected at a single point. The system is consistent. The equations are independent (different graphs), and the system has one solution: (  2, 4).

Slide 9- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7)

Slide 9- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7)

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations by Substitution 9.2 1.Solve systems of linear equations using substitution. 2.Solve applications involving two unknowns using a system of equations.

Slide 9- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Solve systems of linear equations using substitution.

Slide 9- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Two Equations Using Substitution To find the solution of a system of two linear equations using the substitution method: 1. Isolate one of the variables in one of the equations. 2. In the other equation, substitute the expression you found in step 1 for that variable. 3. Solve this new equation. (It will now have only one variable.) 4. Using one of the equations containing both variables, substitute the value you found in step 3 for that variable and solve for the value of the other variable. 5. Check the solution in the original equations.

Slide 9- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. The second equation is solved for x. Step 2: Substitute x = 1 – y for x in the first equation.

Slide 9- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Step 3: Solve for y. Step 4: Solve for x by substituting 4 for y in one of the original equations. The solution is (  3, 4).

Slide 9- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. Use either equation. 2x + y = 7 y = 7 – 2x Step 2: Substitute y = 7 – 2x for y in the second equation.

Slide 9- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Step 3: Solve for x. Step 4: Solve for y by substituting 2 for x in one of the original equations. The solution is (2, 3).

Slide 9- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inconsistent Systems of Equations The system has no solution because the graphs of the equations are parallel lines. You will get a false statement such as 3 = 4. Consistent Systems with Dependent Equations The system has an infinite number of solutions because the graphs of the equations are the same line. You will get a true statement such as 8 = 8.

Slide 9- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations using substitution. Solution Substitute y = 4 – 3x into the second equation. True statement. The number of solutions is infinite.

Slide 9- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve applications involving two unknowns using a system of equations.

Slide 9- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Applications Using a System of Equations To solve a problem with two unknowns using a system of equations: 1. Select a variable for each of the unknowns. 2. Translate each relationship to an equation. 3. Solve the system.

Slide 9- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Terry is designing a table so that the length is twice the width. The perimeter is to be 216 inches. Find the length and width of the table. Understand We are given two relationships and are to find the length and width. Plan Select a variable for the length and another variable for the width, translate the relationships to a system of equations, and then solve the system.

Slide 9- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Execute Let l represent the length and w the width. Relationship 1:The length is twice the width. Translation: l = 2w Relationship 2:The perimeter is 216 inches. Translation: 2l + 2w = 216 System:

Slide 9- 33 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Solve. Now find the value of l. l = 2w = 2(36) = 72 Answer The length should be 72 inches and the width 36 inches. Substitute 2w for l. Combine like terms. Divide both sides by 6 to isolate w.

Slide 9- 34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6)

Slide 9- 35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6)

Slide 9- 36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2)

Slide 9- 37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2)

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations by Elimination 9.3 1.Solve systems of linear equations using elimination. 2.Solve applications using elimination.

Slide 9- 39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Solve systems of linear equations using elimination.

Slide 9- 40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the system of equations. Solution We can add the equations. Notice that y is eliminated, so we can easily solve for the value of x. Divide both sides by 4 to isolate x. Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x + y = 9 4 + y = 9 y = 5 The solution is (4, 5).

Slide 9- 41 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Multiplying Solve the system of equations. Solution Because no variables are eliminated if we add, we rewrite one of the equations so that it has a term that is the additive inverse of one of the terms in the other equation. Multiply the first equation by 4. Solve for y. x + y = 8 3 + y = 8 y = 5 The solution is (3, 5).

Slide 9- 42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Multiplying Solve the system of equations. Solution Choose a variable to eliminate, y, then multiply both equations by numbers that make the y terms additive inverses. Multiply the first equation by 2. Multiply the second equation by  5. Multiply by 2. Multiply by  5. Add the equations to eliminate y.

Slide 9- 43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Substitute x = 1 into either of the original equations. 4x – 5y = 19 4(1) – 5y = 19 4 – 5y = 19 –5y = 15 y =  3 The solution is (1, –3).

Slide 9- 44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Fractions or Decimals Solve the system of equations. Solution To clear the decimals in Equation 1, multiply by 100. To clear the fractions in Equation 2, multiply by 5. Multiply by 100. Multiply by 5. Multiply equation 2 by  1 then combine the equations.

Slide 9- 45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Substitute to find y. The solution is (  1,  3).

Slide 9- 46 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Equations Using Elimination To solve a system of two linear equations using the elimination method: 1. Write the equations in standard form (Ax + By = C). 2. Use the multiplication principle to clear fractions or decimals. 3. Multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. 4. Add the equations. The result should be an equation in terms of one variable. 5. Solve the equation from step 4 for the value of that variable. 6. Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7. Check your solution in the original equations.

Slide 9- 47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inconsistent Systems and Dependent Equations When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent. When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent, so there are an infinite number of solutions.

Slide 9- 48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve applications using elimination.

Slide 9- 49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The length of Richard’s garden is 4 meters greater than 3 times the width. The perimeter of his garden is 72 meters. What are the dimensions of the garden? Understand We are to find the dimensions of the garden. The formula for the perimeter is P = 2l + 2w. Plan Translate the relationships to a system of equations, and then solve the system. Execute Let l = the length Let w = the width

Slide 9- 50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Relationship 1 The length is 4 m more than 3 times the width Translation l = 3w + 4. Relationship 2 The perimeter of the garden is 72 m Translation 2l + 2w = 72. Solve the system: rewrite Multiply by  2.

Slide 9- 51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Now we can find l using l = 3w + 4 l = 3(8) + 4 l = 28 Answer The garden has a length of 28 meters and a width of 8 meters. Check 2l + 2w = 72 2(28) + 2(8) = 72 56 + 16 = 72 True

Slide 9- 52 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5)

Slide 9- 53 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5)

Slide 9- 54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution

Slide 9- 55 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution

Slide 9- 56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution

Slide 9- 57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations in Three Variables 9.4 1.Determine if an ordered triple is a solution for a system of equations. 2.Understand the types of solution sets for systems of three equations. 3.Solve a system of three linear equations using the elimination method.

Slide 9- 59 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Determine whether (2, –1, 3) is a solution of the system Solution In all three equations, replace x with 2, y with –1, and z with 3. x + y + z = 4 2 + (–1) + 3 = 4 4 = 4 TRUE 3 = 3 TRUE 2x – 2y – z = 3 2(2) – 2(–1) – 3 = 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) = –3 –3 = –3 TRUE Because (2,  1, 3) satisfies all three equations in the system, it is a solution for the system.

Slide 9- 60 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve the system using elimination. We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2): (1) (2) (3) (1) (2) (4) 2x + 3y = 8 Adding to eliminate z

Slide 9- 61 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z. (5) x – y + 3z = 8 Multiplying equation (2) by 3 4x + 5y = 14. 3x + 6y – 3z = 6 Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system, 4x + 5y = 14 2x + 3y = 8 (5) (4)

Slide 9- 62 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We multiply both sides of equation (4) by –2 and then add to equation (5): Substituting into either equation (4) or (5) we find that x = 1. 4x + 5y = 14 –4x – 6y = –16, –y = –2 y = 2 Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

Slide 9- 63 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Let’s use equation (1) and substitute our two numbers in it: We have obtained the ordered triple (1, 2, 3). It should check in all three equations. x + y + z = 6 1 + 2 + z = 6 z = 3. The solution is (1, 2, 3).

Slide 9- 64 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three Equations Using Elimination 1. Write each equation in the form Ax + By+ Cz = D. 2. Eliminate one variable from one pair of equations using the elimination method. 3. If necessary, eliminate the same variable from another pair of equations. 4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method. 5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. 6. Check the ordered triple in all three of the original equations.

Slide 9- 65 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve the system using elimination. The equations are in standard form. (1) (2) (3) (2) (3) (4) 3x + 2y = 4 Adding Eliminate z from equations (2) and (3).

Slide 9- 66 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Eliminate z from equations (1) and (2). Multiplying equation (2) by 6 (1) (2) (3) Adding 15x + 15y = 15 Eliminate x from equations (4) and (5). 3x + 2y = 4 15x + 15y = 15 Multiplying top by  5  15x – 10y =  20 15x + 15y = 15 Adding 5y =  5 y =  1 Using y =  1, find x from equation 4 by substituting. 3x + 2y = 4 3x + 2(  1) = 4 x = 2 continued

Slide 9- 67 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Substitute x = 2 and y =  1 to find z. x + y + z = 2 2 – 1 + z = 2 1 + z = 2 z = 1 The solution is the ordered triple (2,  1, 1). (1) (2) (3)

Slide 9- 68 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example At a movie theatre, Kara buys one popcorn, two drinks and 2 candy bars, all for \$12. Rebecca buys two popcorns, three drinks, and one candy bar for \$17. Leah buys one popcorn, one drink and three candy bars for \$11. Find the individual cost of one popcorn, one drink and one candy bar. Understand We have three unknowns and three relationships, and we are to find the cost of each. Plan Select a variable for each unknown, translate the relationship to a system of three equations, and then solve the system.

Slide 9- 69 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Execute: p = popcorn, d = drink, and c = candy Relationship 1: one popcorn, two drinks and two candy bars, cost \$12 Translation: p + 2d + 2c = 12 Relationship 2: two popcorns, three drinks, and one candy bar cost \$17 Translation: 2p + 3d + c = 17 Relationship 3: one popcorn, one drink and three candy bars cost \$11 Translation: p + d + 3c = 11

Slide 9- 71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Use equations 4 and 5 to eliminate d. Substitute for c in d – c = 1 Substitute for c and d in p + d + 3c = 11 p + 2.5 + 3(1.5) = 11 p + 7 = 11 p = 4 The cost of one candy bar is \$1.50. The cost of one drink is \$2.50. The cost of one popcorn is \$4.00.

Slide 9- 72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No

Slide 9- 73 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No

Slide 9- 74 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution

Slide 9- 75 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution

Slide 9- 76 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 90 red wires. How many of each cable were made? a) 10 cable A, 30 cable B, 20 cable C b) 20 cable A, 30 cable B, 10 cable C c) 10 cable A, 103 cable B, 20 cable C d) 10 cable A, 30 cable B, 93 cable C

Slide 9- 77 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 90 red wires. How many of each cable were made? a) 10 cable A, 30 cable B, 20 cable C b) 20 cable A, 30 cable B, 10 cable C c) 10 cable A, 103 cable B, 20 cable C d) 10 cable A, 30 cable B, 93 cable C

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations Using Matrices 9.5 1.Write a system of equations as an augmented matrix. 2.Solve a system of linear equations by transforming its augmented matrix to echelon form.

Slide 9- 79 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Matrix: A rectangular arrow of numbers. The following are examples of matrices: The individual numbers are called elements.

Slide 9- 80 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The rows of a matrix are horizontal, and the columns are vertical. column 1column 2column 3 row 1 row 2 row 3

Slide 9- 81 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Augmented Matrix: A matrix made up of the coefficients and the constant terms of a system. The constant terms are separated from the coefficients by a dashed vertical line. Let’s write this system as an augmented matrix:

Slide 9- 82 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Row Operations The solution of a system is not affected by the following row operations in its augmented matrix: 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied (or divided) by any nonzero real number. 3. Any row may be replaced by a row resulting from adding the elements of that row (or multiples of that row) to a multiple of the elements of any other row.

Slide 9- 83 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Echelon form: An augmented matrix whose coefficient portion has 1s on the diagonal from upper left to lower right and 0s below the 1s.

Slide 9- 84 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Example Solve the following linear system by transforming its augmented matrix into echelon form. We write the augmented matrix. We perform row operations to transform the matrix into echelon form. 3R 2 + R 1

Slide 9- 85 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The resulting matrix represents the system: R 2  10 continued Since y = 1, we can solve for x using substitution. The solution is (2, 1). 3x + 1 = 7 3x + y = 7 3x = 6 x = 2

Slide 9- 86 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Write the augmented matrix. Use the echelon method to solve

Slide 9- 87 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Our goal is to transform into the form Interchange Row 1 and Row 2 –2R 1 + R 2 continued

Slide 9- 88 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley –R 2 + R 3 (1/3)R 2 R 1 + R 3 (1/3)R 3 continued

Slide 9- 89 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley z = –1. Substitute z into y – z = 2 y –(  1) = 2 y + 1 = 2 y = 1 continued Substitute y and z into x – y + z = 1 x – 1 – 1 = 1 x – 2 = 1 x = 3 The solution is (3, 1, –1).

Slide 9- 90 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Replace R 2 in with R 1 + R 2. a)b) c)d)

Slide 9- 91 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Replace R 2 in with R 1 + R 2. a)b) c)d)

Slide 9- 92 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by transforming the augmented matrix into echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution

Slide 9- 93 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by transforming the augmented matrix into echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Equations Using Cramer’s Rule 9.6 1.Evaluate determinants of 2  2 matrices. 2.Evaluate determinants of 3  3 matrices. 3.Solve systems of equations using Cramer’s Rule.

Slide 9- 95 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Square matrix: A matrix that has the same number of rows and columns. Every square matrix has a determinant. Determinant of a 2  2 Matrix If then det(A) =

Slide 9- 96 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the determinant of the following matrix. Solution

Slide 9- 97 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Minor: The determinant of the remaining matrix when the row and column in which the element is located are ignored.

Slide 9- 98 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the minor of  4 in Solution To find the minor of  4, we ignore its row and column (shown in blue) and evaluate the determinant of the remaining matrix (shown in red).

Slide 9- 99 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determinant of a 3  3 Matrix If

Slide 9- 100 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule The solution to the system of linear equations

Slide 9- 101 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The solution to the system of linear equations

Slide 9- 102 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Example Use Cramer’s Rule to solve First, we find D, D x, and D y. The solution is (1/5, 4/5).

Slide 9- 103 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Example Use Cramer’s Rule to solve We need to find D, D x, D y, and D z.

Slide 9- 105 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued The solution is (3, 1, –1). The check is left to the student.

Slide 9- 106 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the determinant. a)  75 b)  72 c)  27 d) 72

Slide 9- 107 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the determinant. a)  75 b)  72 c)  27 d) 72

Slide 9- 108 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve using Cramer’s Rule. a) (  1,  4) b) (1,  3) c) (  2,  3) d) No solution

Slide 9- 109 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve using Cramer’s Rule. a) (  1,  4) b) (1,  3) c) (  2,  3) d) No solution

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Linear Inequalities 9.7 1.Graph the solution set of a system of linear inequalities. 2.Solve applications involving a system of linear inequalities.

Slide 9- 111 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Graph the solution set of a system of linear inequalities.

Slide 9- 112 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving a System of Linear Inequalities To solve a system of linear inequalities, graph all of the inequalities on the same grid. The solution set for the system contains all ordered pairs in the region where the inequalities’ solution sets overlap along with ordered pairs on the portion of any solid line that touches the region of overlap.

Slide 9- 113 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. Both lines will be dashed. Solution

Slide 9- 114 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. Solution

Slide 9- 115 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Inconsistent System Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. The slopes are equal, so the lines are parallel. Since the shaded regions do not overlap there is no solution for the system.

Slide 9- 116 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve applications involving a system of linear inequalities.

Slide 9- 117 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Mr. Reynolds is landscaping his yard. He is looking at some trees and bushes. He would like to purchase at least 3 plants. The trees cost \$40 and the bushes cost \$20. He cannot spend more than \$300 for the plants. Write a system of inequalities that describes what Mr. Reynolds could purchase, then solve the system by graphing. Understand We must translate to a system of inequalities, and then solve the system.

Slide 9- 118 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Plan and Execute Let x represent the trees and y represent the bushes. Relationship 1: Mr. Reynolds would like to purchase at least 3 plants. x + y  3 Relationship 2: Mr. Reynolds cannot spend more than \$300. 40x + 20y  300

Slide 9- 119 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Answer Since Mr. Reynolds cannot purchase negative plants, the solution set is confined to Quadrant 1. Any ordered pair in the overlapping region is a solution. Assuming that only whole trees and bushes can be purchased, only whole numbers would be in the solution set. For example: (4, 2); (3, 4)