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

Prerequisite Check: If you do not know, you need to let me know Simplify Evaluate 3 𝑥−2𝑦 −4(−4𝑥+𝑦) −6(3𝑥+𝑦) 5(4𝑥−3𝑦) What do you have if you have twice of a bag with 2 apples and 3 oranges? What is twice of 2𝑥+3𝑦? What is −6 times 𝑥−2𝑦?

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

Solve the linear system using the linear combination method. Example 1 Multiply One Equation Solve the linear system using the linear combination method. 6 3y 2x = – Equation 1 8 5y 4x = – Equation 2 SOLUTION STEP 1 Multiply the first equation by 2 so that the coefficients of x differ only in sign. – 6 3y 2x = – 8 5y 4x 12 6y + 4 y = – 5

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

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

Solve the system using the linear combination method. Example 2 Multiply Both Equations Solve the system using the linear combination method. 22 12y 7x = – Equation 1 14 8y 5x = + – Equation 2 SOLUTION STEP 1 Multiply the first equation by 2 and the second equation by 3. 22 12y 7x = – 14 8y 5x + 44 24y 14x 42 15x 2 x = – 8

Add the revised equations and solve for x. Example 2 Multiply Both Equations STEP 2 Add the revised equations and solve for x. 2 x = – 2 x = STEP 3 Substitute 2 for x in one of the original equations and solve for y. 14 8y 5x = + – Write Equation 2. 14 8y = + Substitute 2 for x. – ( ) 2 5 14 8y 10 = + – Multiply. 3 y = Solve for y. 9

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

Solve the system using the linear combination method. Example 3 A Linear System with No Solution Solve the system using the linear combination method. 12 8y 4x = + – Equation 1 7 4y 2x = – Equation 2 SOLUTION Multiply the second equation by 2 so that the coefficients of y differ only in sign. 8y – 12 4x = + 14 7 4y 2x 2 = Add the revised equations. 11

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

Solve the system using the linear combination method. Checkpoint Solve a Linear System Solve the system using the linear combination method. 1. 5 4y x = – ANSWER (1, 1) – 2x + y = 1 2. 4 y 2x = – 8 2y 4x ANSWER infinitely many solutions 3. 2 2y 3x = – 1 3y 4x ANSWER (4, 5 )

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

Use a Linear System as a Model Example 4 Use a Linear System as a Model 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?

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

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

Use a Linear System as a Model Example 4 Use a Linear System as a Model ALGEBRAIC MODEL 30 = 10p + 5s Equation 1 (servings needed) 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 + 5s 80 40p 10s 20p – 60 20 = 20p Add the revised equations and solve for p. 1 = p

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

2 pans of pasta and 4 sandwich trays 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

Conclusions Pg142 #(10,14,26,28,34) Due by the end of the class. 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.

Choosing a Method What method is more convenient? Graphing, Substitution, or Linear Combination.

Choosing a Method 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 Which Method Will You Choose? Substitution Method: Step-by-Step Choose a method to solve the linear system. Explain your choice. Then solve the system. −3𝑥+𝑦=−5 5𝑥−3𝑦=3 Solve one equation for one of its variables Substitute the expression from Step 1 into the other equation and solve for the other variable Substitute the value from Step 2 into the revised equation from Step 1 and solve Check the solution in each of the original equations

Choosing a Method Which Method Will You Choose? Linear Combination Method: Step-by-Step Choose a method to solve the linear system. Explain your choice. Then solve the system. 2𝑥−5𝑦=−1 4𝑥+3𝑦=11 Multiply, if necessary, one or both equations by a constant so that the coefficients of one of the variables differ only in sign. Add the revised equations from Step 1. combining like terms will eliminate one variable. Solve for the remaining variable. Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable. Check the solution in each of the original equations.

Choosing a Method Choose a method to solve the linear system. Explain your choice. Then solve the system. 2𝑥+𝑦=7 3𝑥+4𝑦=8 Answer: Substitution Method y has a coefficient of 1 (4,−1)

Choosing a Method Choose a method to solve the linear system. Explain your choice. Then solve the system. 3𝑥+5𝑦=−1 6𝑥+7𝑦=−5 Answer: Linear Combination Method Neither variable has a coefficient of 1 or -1 (−2,1)

Choosing a Method Choose a method to solve the linear system. Explain your choice. Then solve the system. 2𝑥−𝑦=7 𝑥+3𝑦=14 Answer: Substitution Method y has a coefficient of -1 and x has a coefficient of 1 (5,3)

Additional Practice Problems solve graphically and algebraically http://www.classzone.com/cz/books/algebra_2_cs/resources/applications/animations/html/explore_learning/chapter_3/dswmedia/7_5_special_sys.html