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## Solving Linear Systems by Linear Combinations

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**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 What do you have if you have twice of a bag with 2 apples and 3 oranges? What is twice of ? What is times ?**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**Example 1**2x – 3y 6 = 4x – 5y 8 = SOLUTION STEP 1 Multiply the first equation by 2 so that the coefficients of x differ only in sign. – – 4x + 6y – 12 = 2x – 3y 6 = 4x – 5y 8 4x – 5y 8 = = y – 4 = Multiply One Equation Solve the linear system using the linear combination method. Equation 1 Equation 2**Example 1**STEP 3 Substitute4 for y in one of the original equations and solve for x. – Write Equation 1. 2x – 3y 6 = ( ) – Substitute 4 for y. 3 4 2x – 6 = – Simplify. 2x + 12 6 = Subtract 12 from each side. – 2x 6 = – Solve for x. x 3 = y – 4 = Multiply One Equation STEP 2 Add the revised equations and solve for y.**Example 1**The solution is . ANSWER ( ) – – 3, 4 STEP 4 Check by substituting 3 for x and 4 for y in the original equations. – – Multiply One Equation**Example 2**– – 7x 12y 22 = – 5x + 8y 14 = SOLUTION STEP 1 Multiply the first equation by 2 and the second equation by 3. – – 14x 24y 44 = – 15x + 24y 42 = – x – 2 = – 5x + 8y 14 – – 7x 12y = 22 = Multiply Both Equations Solve the system using the linear combination method. Equation 1 Equation 2**Example 2**– x – 2 = x 2 = STEP 3 Substitute2 for x in one of the original equations and solve for y. Write Equation 2. ( ) – 5 2 Substitute 2 for x. + 8y 14 = – Multiply. 10 + 8y 14 = Solve for y. y 3 = – 5x + 8y 14 = Multiply Both Equations STEP 2 Add the revised equations and solve for x.**Example 2**The solution is (2, 3). ANSWER Multiply Both Equations STEP 4 Check by substituting 2 for x and 3 for y in the original equations.**Example 3**2x – 4y 7 = SOLUTION – 4x + 8y – 12 = Multiply the second equation by 2 so that the coefficients of y differ only in sign. – 4x + 8y – 12 – 4x + 8y – 12 = = 2x – 4y 7 4x – 8y 14 = = Add the revised equations. 0 2 = A Linear System with No Solution Solve the system using the linear combination method. Equation 1 Equation 2**Example 3**A Linear System with No Solution ANSWER Because the statement 02 is false, there is no solution. =**Checkpoint**(1, 1) – ANSWER x – 4y 5 = 2. 2x – y 4 = infinitely many solutions ANSWER 4x – 2y 8 = 3. 3x – 2y 2 = (4, 5 ) ANSWER 4x – 3y 1 = Solve a Linear System Solve the system using the linear combination method. 1. 2x + y 1 =**Checkpoint**ANSWER if you get a false equation; if you get a true equation Solve a Linear System 4. How can you tell when a system has no solution? infinitely many solutions?**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?**Servings per**sandwich tray Servings per pan Sandwich trays Servings needed Money to spend on food Price per pan Price per tray Sandwich trays Example 4 Use a Linear System as a Model SOLUTION VERBAL MODEL Pans of pasta + • • = Pans of pasta + • • =**Example 4**Use a Linear System as a Model (servings) Servings per pan of pasta 10 LABELS = (pans) Pans of pasta p = (servings) Servings per sandwich tray 5 = (trays) Sandwich trays s = (servings) Servings needed 30 = (dollars) Price per pan of pasta 40 = (dollars) Price per sandwich tray 10 = (dollars) Money to spend on food 80 =**+**+ 10p 10p 5s 5s 30 30 = = + + Equation 2 (money to spend on food) 40p 40p 10s 10s 80 80 = = Multiply Equation 1 by 2 so that the coefficients of s differ only in sign. – – – – 20p 10s 60 = + 40p 10s 80 = 20p 20 = Add the revised equations and solve for p. p 1 = Example 4 Use a Linear System as a Model ALGEBRAIC MODEL Equation 1 (servings needed)**+**10p 5s 30 = Write Equation 1. ( ) 1 Substitute 1 for p. + 10 5s 30 = Subtract 10 from each side. 5s 20 = Solve for s. s 4 = ANSWER The caterer should make 1 pan of pasta and 4 sandwich trays. Example 4 Use a Linear System as a Model Substitute 1 for p in one of the original equations and solve for s.**Checkpoint**ANSWER 2 pans of pasta and 4 sandwich trays Solve a Linear System 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? 5.**Conclusions**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. 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. 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. Answer: Substitution Method y has a coefficient of 1**Choosing a Method**Choose a method to solve the linear system. Explain your choice. Then solve the system. Answer: Linear Combination Method Neither variable has a coefficient of 1 or -1**Choosing a Method**Choose a method to solve the linear system. Explain your choice. Then solve the system. Answer: Substitution Method y has a coefficient of -1 and x has a coefficient of 1**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