Ch 3 Alg Note Sheet.doc 3.1 Graphing Sstems of Equations Sstems of Linear Equations A sstem of equations is a set of two or more equations that use the same variables. If the graph of each equation =.4 5 in a sstem of two variables is a line, then the sstem is a linear sstem. For eample:. = + 3 A solution of a sstem of equations is a set of values for the variables that makes all the equations true. You can solve some linear sstems b graphing the equations. The points where both (or all) the graphs intersect represent the solutions. Eample 1 Solving b Graphing Graph + = 7 Graph 3 = 3 6 4 4 4 6 Common Solution: Check solution + = 7 Check solution 3 = Eample a Solve = + 5 = + 6 4 6 4 4 6 8 4 6 8 Compare the slopes and - intercepts. How man solutions? Eample b Solve = 3 + = 3 1 Compare the slopes and - intercepts. How man solutions? 6 4 6 4 4 6 8 4 6 8 Eample c Solve = 4 8 + 4 Rewrite! = 6 4 4 6 8 4 6 8 How man solutions? 6 4 S. Stirling Page 1 of 1
Ch 3 Alg Note Sheet.doc Summarize Eample : Description of lines. a. b. c. coinciding How Man Points of Intersection? Equal slopes? Classifing Sstems of Linear Equations Same - intercept? Since ou can determine the number of points of intersection (the number of solutions b the slope and -intercept, do ou reall need to graph them to determine the number of solutions? Or, could ou simpl eamine the slopes and -intercepts? You just need to get the lines in slope-intercept form. You can classif a sstem of two linear equations b the number of solutions. An consistent is a sstem that has at least one solution. A sstem that has a unique (onl one) solution, as in E 1 and E a, is an independent sstem. However, not ever sstem has a unique solution. A dependent does not have a unique solution, it has an infinite number of solutions, E c (shares all points). An inconsistent is a sstem that has no solution, E b (shares no points). Eample 3 Classifing Without Graphing Classif the sstems without graphing.. Eample 3a = + 3 + = 1 Eample 3b = + 3 4 + = 6 Eample 3c = 5 + 3 = S. Stirling Page of 1
Ch 3 Alg Note Sheet.doc 3. Solving Sstems Algebraicall Wh other methods? Not ever sstem can be solved easil b graphing. The eact point of intersection is not obvious. Algebraic methods allow ou to find eact solutions without using a graphing, which can be tedious. Part A: Solving Sstems b Elimination Remember, the main Properties of Equalit for solving equations? What ever ou do to one side ou must do to the other. This technique for solving sstems is based on adding (or subtracting) the same quantit to both sides of an equation. The strange part is that even though the quantities are equal, the don t look the same, but the are equal! Also, we won t be able to find a unique solution if we can t isolate a variable. Can we add the same quantit to both sides so that one of the variables goes awa? The trick to this method is to make a pair of like terms additive inverses so that when ou add the equations together, ou eliminate a term. a + ( a) = = Eample 1 Solving b Elimination 4 = 7 Solve the sstem b elimination. + = 3 Since + and 3 are equal, ou can add each to both sides of 4 = 7. Note: the -terms are additive inverses, + = =, so ou successfull eliminate a variable. Eample Solving b Elimination 4 = 18 Solve the sstem b elimination. + 3 = 11 Step 1: 4 = 7 + = 3 5 = 1 = Step : Find matching -value: + = 3 () + = 3 = 1 = 1, 1 Solution ( ) Eample 3 Solving b Elimination 3 + 4 =.5 Solve the sstem b elimination. 5 4 = 1.5 It is not alwas that eas! You ma need to make two terms additive inverses b multipling one or both equations in the sstem b a nonzero number. (Properties of Equalit again!) In doing so, ou create a sstem equivalent to the original one. Equivalent sstems are sstems that have the same solution(s). Remember to full use the distributive propert (ever term gets multiplied b number) along with the multiplication propert of equalit (both sides get multiplied b the same number). S. Stirling Page 3 of 1
Ch 3 Alg Note Sheet.doc Eample 4 Solving b Elimination, Equivalent Sstems = 11 Solve the sstem b elimination. 3 3 = 3 Adding the same thing to both sides will not eliminate a variable in this case, so to eliminate the -terms, make the coefficients of 3 and 3. Eample 4 Alternate Process Divide both sides of the nd equation b 3. = 11 multipl equation b 3 3 3 = 3 leave the nd equation alone Now add the equations Eample 5 Solving b Elimination, Equivalent Sstems 3 + 7 = 15 Solve the sstem b elimination. 5 + = 4 To eliminate the -terms, make the coefficients of 14 and 14. Eample 5 Alternate Process If ou choose to eliminate the -terms instead, make the coefficients 15 and 15. 3 + 7 = 15 multipl equation b 5 + = 4 multipl equation b 7 Now add the equations S. Stirling Page 4 of 1
Ch 3 Alg Note Sheet.doc Part B: Solving Sstems b Substitution The main idea? Since the solution is where the two lines come together, is there a wa to put the two equations that represent the lines together? Can ou solve one of the equations for one of the variables and then put it into the other equation? Bring them together? Also, remember that a main propert in Algebra sas that ou ma substitute one quantit for another if the represent the same quantit (are equal). Eample 1 Solving b Substitution 4 + 3 = 4 Solve the sstem b substitution. = 7 Solve equation # for, because it s eas to do! = 7 = 7 Substitute in for in equation #1. You bring them together. Intersect where 1 =. 4 + 3 = 4 4 + 3( 7) = 4 4 + 6 1 = 4 1 1 1 = 4 1 = 5 =.5 Now ou can find a matching for our. Does it matter which equation ou use? Choose the easiest to find the matching - value solve, and use the other to check. = 7 = (.5) 7 = Since our solution is a point (an ordered pair), ou should write it that wa!.5, Solution to the sstem: ( ) Check: 4 + 3 = 4 4(.5) + 3( ) = 4 1 6 = 4 Eample 1a Solve the sstem b substitution. 3 = 6 + = 1 Eample 1b 3 = Solve the sstem b substitution. 4 + 3 = 6 S. Stirling Page 5 of 1
Ch 3 Alg Note Sheet.doc Part C: Solving Sstems Applications and Unusual Solutions Eample 6 Solving Real Problems Refer to the advertisement at the right. The cost of membership in a health club includes a monthl charge and a one-time initiation fee. Find the monthl charge and the initiation fee. Review our notes on the tpes of linear sstems: consistent (independent and dependent) and inconsistent. Before, we determined the tpe based on graphing and determining the number of solutions. What happens if we use algebra techniques to tr to solve the sstem? How will we know if the sstem is dependent or inconsistent? Eample 7 Solving a Sstem Without a Unique Solution Don t panic if both of the variables are eliminated. There is an interpretation for this. 5a. = 3 + = 3 5b. 3 = 18 + 3 = 6 Eliminate the -terms: = 3 + = 3 = Interpret: Get an equation that is alwas TRUE. The sstem has an infinite number of solutions. The sstem is dependent. The sstem represents the same line. Solution: All real numbers where = 3 or {(, ) = 3} Eliminate the -terms: 3 = 18 + 3 = 6 = 1 Interpret: Get an equation that is alwas FALSE. The sstem has no solution. The sstem is inconsistent. The sstem represents parallel lines. Solution: No Solution S. Stirling Page 6 of 1
Ch 3 Alg Note Sheet.doc 3.3 Sstems of Inequalities Solving Sstems of Inequalities You can solve a sstem of inequalities b creating a table and testing values in the inequalities. Not ver efficient. and there is an infinite number of solutions. A better wa to solve a sstem is b graphing. You can solve a sstem of linear inequalities b graphing. Recall from Lesson -7 that when the variables represent real numbers, the solutions of an inequalit include all the points on one side of a boundar line. Thus, for two inequalities, ever point in the region of overlap of the two solutions is a solution of the sstem. Eample Solving a Sstem b Graphing Solve the sstem of inequalities. Graph + = 6 + < 6 3 + 5 3 Graph = + 5 6 4 4 4 6 8 4 6 Eample 3 Application An entrance eam has two parts, a verbal part and a mathematics part. You can score a maimum total of 16 points. For admission, the school of our choice requires a math score of at least 6. Write a sstem of inequalities to model scores that meet the school s requirements. Then solve the sstem. 15 Math Score 1 5 5 1 15 Verbal Score S. Stirling Page 7 of 1
Ch 3 Alg Note Sheet.doc Eample 4 Solving a Linear Absolute Value Sstem Solve the sstem of inequalities. Graph < 4 < 4 3 Graph 3 4 4 4 6 4 6 Sometimes graphing is not necessar. The book uses a table method, however here an analtical method is easier. Eample 1 Solving a Sstem (Alternate Method) A science class has 6 computers for students. Students have the option of using a computer program to investigate frog biolog or using a computer and their graphing calculators to investigate the properties of heat transfer. Each frog lab must have 3 students in a group. Each heat lab must have 4 students in a group. In how man was can ou set up the lab groups? Let h = number of heat lab groups Let f = number of frog lab groups Set up inequalities using the information given: # lab groups less than or equal to 6: h + f 6 # students equal to : 3 f + 4h = The situation is discrete. h and f must be whole numbers. Use substitution to solve: 4h = 3 f 3 h = 5 f Now substitute into the inequalit 4 3 5 6 4 f + f 1 1 4 f and f 4 Test possible whole number values: f = 4 3(4) + 4h = 4h = 8 h = 4 frog groups & heat groups f = 3 3(3) + 4h =, 4h = 11 f = 3() + 4h =, 4h = 14 f = 1 3(1) + 4h =, 4h = 17 All ield numbers that are not a whole numbers But if f = 3() + 4h = 4h = h = 5 frog groups & 5 heat groups S. Stirling Page 8 of 1
Ch 3 Alg Note Sheet.doc 3.4 Linear Programming Use the Activit: Finding Maimum and Minimum Values to help ou comprehend the vocabular and process summarized below. After finishing the activit, label the sections with the appropriate vocabular. Linear Programming is a technique that identifies the minimum or maimum value of some quantit. This quantit is modeled with an objective function. In the activit, the cost equation was the objective function, C = 8 + 1. The limits ou have on the variables in the given situation are called constraints. The are written as linear inequalities. In the activit, the inequalities ou wrote from the situation were the constraints on the variables. The region that satisfies all the constraints is called the feasible region. This is the solution set defined b the graph of the constraints. The region can be man different shapes, be prepared. Graphs of an objective function that represent a maimum or minimum value (is a line) and it will alwas intersect a feasible region at a verte, the corners of the feasible region. The maimum or minimum value will alwas occur at a verte. Eample 1 Testing Vertices What values of and maimize P for the objective function P = 3 +? Constraints: 3 3 + 7, When = 4 and = 3, P has its maimum value of 18. The answer is (4, 3). Eample 1 (cont.) Use the constraints in Eample 1 with the objective function P = + 3. Find the values of and that maimize and minimize P. Find the value of P at each point. S. Stirling Page 9 of 1
Ch 3 Alg Note Sheet.doc Linear Programming General Procedure Linear Programming is a technique that identifies the minimum or maimum value of some quantit. Limits on the variables in the objective function are constraints written as linear inequalities. The region that satisfies all the constraints is called the feasible region. Each corner of the feasible region is called a verte. The quantit that ou want to minimize or maimize is modeled with an objective function (equation). You will follow the same procedure on all problems: 1. Identif and carefull define the variables. Ask ourself what quantities can change in the problem? Use letters that make sense in the problem.. Identif the constraints. Ask ourself what are the limits on the variables? Organize this information into a table then write the inequalities that represent the constraints. 3. Write the objective function. Ask ourself what am I tring to maimize or minimize? Find the information in the problem that will help ou calculate that quantit. And write an equation. 4. Graph the feasible region. Find out where the inequalities (constraints) overlap. An combination of quantities (the points) in this region meets all of the constraints. You will find that graphing b - and - intercepts will help ou the most! Use the intercepts to plan the scales on each ais. Should ou count b 1,, 5s? (The scales on each ais should be the same, if possible, to make the graph eas to interpret.) Tr to fill as much of the grid as possible. Make the graph as big as possible. Label the aes in words!! 5. Find the vertices, corner points, of the feasible region. First label them so ou remember what ou are finding: points A, B, C Use the appropriate methods (substitution and elimination) for finding these because ou can t count on the graph to be accurate. You can use our graph to see if our answers make sense. 6. Test the vertices (corner points) in the objective equation. Find the combination of quantities (the point) that will either maimize or minimize the objective function. 7. Summarize our results in words. S. Stirling Page 1 of 1
Ch 3 Alg Note Sheet.doc Eample Suppose ou are selling cases of mied nuts and roasted peanuts. You can order no more than a total of 5 cans and packages and spend no more than $6. How can ou maimize our profit? How much is the maimum profit? 1. Define Variables:. Write the Constraints: Organize the information in a table. Write the constraints. Write the objective function. Constraints Mied Nuts Roasted Peanuts From the tet As an Inequalit The Objective is? 3. Write the Objective Function: 4. Graph the Feasible Region: (- and -intercepts work well!) 1 S. Stirling Page 11 of 1
Ch 3 Alg Note Sheet.doc 5. Find the Vertices: (Solve sstems and or find intercepts. Write as points.) 6. Test the Vertices: Evaluate the corner points in the objective function. 7. Summarize Results S. Stirling Page 1 of 1