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AII, 2.0: STUDENTS SOLVE SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES (IN TWO OR THREE VARIABLES) BY SUBSTITUTION, WITH GRAPHS, OR WITH MATRICES. LA, 6.0:

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Presentation on theme: "AII, 2.0: STUDENTS SOLVE SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES (IN TWO OR THREE VARIABLES) BY SUBSTITUTION, WITH GRAPHS, OR WITH MATRICES. LA, 6.0:"— Presentation transcript:

1 AII, 2.0: STUDENTS SOLVE SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES (IN TWO OR THREE VARIABLES) BY SUBSTITUTION, WITH GRAPHS, OR WITH MATRICES. LA, 6.0: STUDENTS DEMONSTRATE AN UNDERSTANDING THAT LINEAR SYSTEMS ARE INCONSISTENT (HAVE NO SOLUTIONS), HAVE EXACTLY ONE SOLUTION, OR HAVE INFINITELY MANY SOLUTIONS Solving Linear Systems by Linear Combinations

2 Objectives Key Words Solve a system of linear equations in two variables by the linear combination method EC: Choosing a Method Linear combination method Solving Linear Systems by Linear Combinations

3 Simplify Evaluate Prerequisite Check: If you do not know, you need to let me know

4 Using the Linear Combination Method Step-by-Step Steps: 1. Multiply, if necessary, one or both equations by a constant so that the coefficients of one of the variables differ only in sign. 2. Add the revised equations from Step 1. combining like terms will eliminate one variable. Solve for the remaining variable. 3. Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable. 4. Check the solution in each of the original equations.

5 Example 1 Multiply One Equation Solve the linear system using the linear combination method. Equation 1 63y3y2x2x = – Equation 2 85y5y4x4x = – SOLUTION STEP 1Multiply the first equation by 2 so that the coefficients of x differ only in sign. – 63y3y2x2x = – 85y5y4x4x = –85y5y4x4x = – 126y6y4x4x = +–– 4y = –

6 Example 1 Multiply One Equation STEP 2Add the revised equations and solve for y. STEP 3Substitute 4 for y in one of the original equations and solve for x. – Write Equation 1. 63y3y2x2x = – Substitute 4 for y. 62x2x = – () 4 – 3 – Simplify. 6122x2x = + Subtract 12 from each side. 62x2x = – Solve for x. 3x = – 4y = –

7 Example 1 Multiply One Equation STEP 4Check by substituting 3 for x and 4 for y in the original equations. –– ANSWER The solution is. () 3,3, –– 4

8 Example 2 Multiply Both Equations Solve the system using the linear combination method. Equation 1 Equation 2 SOLUTION STEP 1Multiply the first equation by 2 and the second equation by y7x7x = –– 148y8y5x5x = +– 2212y7x7x = –– 148y8y5x5x = + – 4424y14x = –– 4224y15x = + – 2x = ––

9 Example 2 Multiply Both Equations STEP 2Add the revised equations and solve for x. 2x = 2x = –– STEP 3Substitute 2 for x in one of the original equations and solve for y. 148y8y5x5x = + – Write Equation y8y = + Substitute 2 for x. – () y8y10 = + – Multiply. 3y = Solve for y.

10 Example 2 Multiply Both Equations STEP 4Check by substituting 2 for x and 3 for y in the original equations. ANSWER The solution is (2, 3).

11 Example 3 A Linear System with No Solution Solve the system using the linear combination method. Equation 1 Equation 2 74y4y2x2x = – 128y8y4x4x = +–– SOLUTION Multiply the second equation by 2 so that the coefficients of y differ only in sign. 8y8y – 124x4x = + 148y8y4x4x = –– 74y4y2x2x = – 128y8y4x4x = +–– 20 = Add the revised equations.

12 Example 3 A Linear System with No Solution ANSWER Because the statement 0 2 is false, there is no solution. =

13 Checkpoint ANSWER infinitely many solutions 1. Solve the system using the linear combination method. Solve a Linear System 54y4yx = – 1y2x2x = + ANSWER (1, 1) – 2. 4y2x2x = – 82y2y4x4x = – 3. 22y2y3x3x = – 13y3y4x4x = – ANSWER (4, 5 )

14 Checkpoint ANSWER if you get a false equation; if you get a true equation 4. How can you tell when a system has no solution? infinitely many solutions? Solve a Linear System

15 Use a Linear System as a Model Example 4 Catering A customer hires a caterer to prepare food for a party of 30 people. The customer has $80 to spend on food and would like there to be a choice of sandwiches and pasta. A $40 pan of pasta contains 10 servings, and a $10 sandwich tray contains 5 servings. The caterer must prepare enough food so that each person receives one serving of either food. How many pans of pasta and how many sandwich trays should the caterer prepare?

16 Use a Linear System as a Model Example 4 SOLUTION VERBAL MODEL Servings per pan Pans of pasta Sandwich trays = + Servings per sandwich tray Servings needed Price per pan Pans of pasta Sandwich trays = + Price per tray Money to spend on food

17 Use a Linear System as a Model Example 4 LABELS Servings per pan of pasta 10 = (servings) Pans of pasta p = (pans) Servings per sandwich tray 5 = (trays) Sandwich trays s = Servings needed 30 = Price per pan of pasta 40 = (dollars) Price per sandwich tray 10 = (dollars) Money to spend on food 80 = (dollars) (servings)

18 Use a Linear System as a Model Example 4 ALGEBRAIC MODEL Equation 1 (servings needed) 30 = 10p + 5s5s Equation 2 (money to spend on food) 80 = 40p + 10s Multiply Equation 1 by 2 so that the coefficients of s differ only in sign. – 30 = 10p + 5s5s 80 = 40p + 10s = 20p10s –– 60 – 80 = 40p + 10s 20 = 20p Add the revised equations and solve for p. 1 = p

19 Use a Linear System as a Model Example 4 ANSWER The caterer should make 1 pan of pasta and 4 sandwich trays. Substitute 1 for p in one of the original equations and solve for s. Write Equation = 10p + 5s5s Substitute 1 for p. 30 = s5s () 1 Subtract 10 from each side. 20 = 5s5s 4 = s Solve for s.

20 Checkpoint Solve a Linear System 5. Another customer asks the caterer in Example 4 to plan a party for 40 people. This customer also wants both sandwiches and pasta and has $120 to spend. How many pans of pasta and how many sandwich trays should the caterer prepare? ANSWER 2 pans of pasta and 4 sandwich trays

21 Summary Assignment How do you solve a system of linear equations algebraically?  To use the linear combination method, multiply one or both equations by constants to get opposite coefficients for one variable. Add the revised equations to solve for one variable. Then substitute the value you found into either one of the original equations to find the value of the other variable. Pg142 #(10,14,26,28,34) Due by the end of the class. Conclusions

22 WHAT METHOD IS MORE CONVENIENT? GRAPHING, SUBSTITUTION, OR LINEAR COMBINATION. Choosing a Method

23 Substitution Linear Combination If one of the variables has a coefficient of 1 or - 1, the substitution method is convenient. In general, you should solve for the variable. If neither variable has a coefficient of 1 or -1, the linear combination method is often more convenient, although you can still use substitution. Choosing a Method

24 Which Method Will You Choose? Substitution Method: Step-by-Step Substitution Method: Step-by-Step 1. Solve one equation for one of its variables 2. Substitute the expression from Step 1 into the other equation and solve for the other variable 3. Substitute the value from Step 2 into the revised equation from Step 1 and solve 4. Check the solution in each of the original equations Choosing a Method

25 Which Method Will You Choose? Linear Combination Method: Step-by-Step Linear Combination Method: Step-by-Step 1. Multiply, if necessary, one or both equations by a constant so that the coefficients of one of the variables differ only in sign. 2. Add the revised equations from Step 1. combining like terms will eliminate one variable. Solve for the remaining variable. 3. Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable. 4. Check the solution in each of the original equations. Choosing a Method

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29 Additional Practice Problems solve graphically and algebraically mations/html/explore_learning/chapter_3/dswmedia/7_5_special_sys.html


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