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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 3-6 Solving Linear Systems in Three Variables Holt Algebra 2 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Algebra 2

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3-6 Solving Linear Systems in Three Variables Warm Up Solve each system of equations algebraically. Classify each system and determine the number of solutions. 1. 2. x = 4y + 10 4x + 2y = 4 6x – 5y = 9 2x – y =1 (2, –2) (–1,–3) 3. 4. 3x – y = 8 6x – 2y = 2 x = 3y – 1 6x – 12y = –4 inconsistent; noneconsistent, independent; one

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Represent solutions to systems of equations in three dimensions graphically. Solve systems of equations in three dimensions algebraically. Objectives

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Systems of three equations with three variables are often called 3-by-3 systems. In general, to find a single solution to any system of equations, you need as many equations as you have variables.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Recall from Lesson 3-5 that the graph of a linear equation in three variables is a plane. When you graph a system of three linear equations in three dimensions, the result is three planes that may or may not intersect. The solution to the system is the set of points where all three planes intersect. These systems may have one, infinitely many, or no solution.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned in Lesson 3-2.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Use elimination to solve the system of equations. Example 1: Solving a Linear System in Three Variables Step 1 Eliminate one variable. 5x – 2y – 3z = –7 2x – 3y + z = –16 3x + 4y – 2z = 7 In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1 and z is easy to eliminate from the other equations. 1 2 3

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Example 1 Continued 5x – 2y – 3z = –7 11x – 11y = –55 3(2x –3y + z = –16) 5x – 2y – 3z = –7 6x – 9y + 3z = –48 1 2 1 4 3x + 4y – 2z = 7 7x – 2y = –25 2(2x –3y + z = –16) 3x + 4y – 2z = 7 4x – 6y + 2z = –32 3 2 Multiply equation - by 3, and add to equation. 1 2 Multiply equation - by 2, and add to equation. 3 2 5 Use equations and to create a second equation in x and y. 3 2

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 11x – 11y = –55 7x – 2y = –25 You now have a 2-by-2 system. 4 5 Example 1 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables –2(11x – 11y = –55) 55x = –165 11(7x – 2y = –25) –22x + 22y = 110 77x – 22y = –275 4 5 1 1 Multiply equation - by –2, and equation - by 11 and add. 4 5 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. x = –3 Solve for x. Example 1 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 11x – 11y = –55 11(–3) – 11y = –55 4 1 1 Step 3 Use one of the equations in your 2-by-2 system to solve for y. y = 2 Substitute –3 for x. Solve for y. Example 1 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 2x – 3y + z = –16 2(–3) – 3(2) + z = –16 2 1 1 Step 4 Substitute for x and y in one of the original equations to solve for z. z = –4 Substitute –3 for x and 2 for y. Solve for y. The solution is (–3, 2, –4). Example 1 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Use elimination to solve the system of equations. Step 1 Eliminate one variable. –x + y + 2z = 7 2x + 3y + z = 1 –3x – 4y + z = 4 1 2 3 Check It Out! Example 1 In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables –x + y + 2z = 7 –5x – 5y = 5 –2(2x + 3y + z = 1) –4x – 6y – 2z = –2 1 2 1 4 5x + 9y = –1 –2(–3x – 4y + z = 4) –x + y + 2z = 7 6x + 8y – 2z = –8 1 3 Multiply equation - by –2, and add to equation. 1 2 1 3 5 Check It Out! Example 1 Continued –x + y + 2z = 7 Use equations and to create a second equation in x and y. 1 3

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables You now have a 2-by-2 system. Check It Out! Example 1 Continued 4 5 –5x – 5y = 5 5x + 9y = –1

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 4y = 4 4 5 1 Add equation to equation. 4 5 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from Lesson 3-2. Solve for y. Check It Out! Example 1 Continued –5x – 5y = 5 5x + 9y = –1 y = 1

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables –5x – 5(1) = 5 4 1 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. x = –2 Substitute 1 for y. Solve for x. Check It Out! Example 1 –5x – 5y = 5 –5x – 5 = 5 –5x = 10

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 2(–2) +3(1) + z = 1 2x +3y + z = 1 2 1 1 Step 4 Substitute for x and y in one of the original equations to solve for z. z = 2 Substitute –2 for x and 1 for y. Solve for z. The solution is (–2, 1, 2). Check It Out! Example 1 –4 + 3 + z = 1

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables You can also use substitution to solve a 3-by-3 system. Again, the first step is to reduce the 3-by-3 system to a 2-by-2 system.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables The table shows the number of each type of ticket sold and the total sales amount for each night of the school play. Find the price of each type of ticket. Example 2: Business Application OrchestraMezzanineBalconyTotal Sales Fri2003040$1470 Sat2506050$1950 Sun150300$1050

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Example 2 Continued Step 1 Let x represent the price of an orchestra seat, y represent the price of a mezzanine seat, and z represent the present of a balcony seat. Write a system of equations to represent the data in the table. 200x + 30y + 40z = 1470 250x + 60y + 50z = 1950 150x + 30y = 1050 1 2 3 Fridays sales. Saturdays sales. Sundays sales. A variable is missing in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 5(200x + 30y + 40z = 1470) –4(250x + 60y + 50z = 1950) 1 Step 2 Eliminate z. Multiply equation by 5 and equation by –4 and add. 12 2 1000x + 150y + 200z = 7350 –1000x – 240y – 200z = –7800 y = 5 Example 2 Continued By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 150x + 30y = 1050 150x + 30(5) = 1050 3 Substitute 5 for y. x = 6 Solve for x. Step 3 Use equation to solve for x. 3 Example 2 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 200x + 30y + 40z = 1470 1 Substitute 6 for x and 5 for y. 1 z = 3 Solve for x. Step 4 Use equations or to solve for z. 21 200(6) + 30(5) + 40z = 1470 The solution to the system is (6, 5, 3). So, the cost of an orchestra seat is $6, the cost of a mezzanine seat is $5, and the cost of a balcony seat is $3. Example 2 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Jadas chili won first place at the winter fair. The table shows the results of the voting. How many points are first-, second-, and third-place votes worth? Name 1st Place 2nd Place 3rd Place Total Points Jada 31415 Maria 24014 Al 22313 Winter Fair Chili Cook-off

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Continued Step 1 Let x represent first-place points, y represent second-place points, and z represent third- place points. Write a system of equations to represent the data in the table. 3x + y + 4z = 15 2x + 4y = 14 2x + 2y + 3z = 13 1 2 3 Jadas points. Marias points. Als points. A variable is missing in one equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 3(3x + y + 4z = 15) –4(2x + 2y + 3z = 13) 1 Step 2 Eliminate z. Multiply equation by 3 and equation by –4 and add. 31 3 9x + 3y + 12z = 45 –8x – 8y – 12z = –52 x – 5y = –7 4 Check It Out! Example 2 Continued 2 –2(x – 5y = –7) 4 2x + 4y = 14 –2x + 10y = 14 2x + 4y = 14 y = 2 Multiply equation by –2 and add to equation. 2 4 Solve for y.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 2x + 4y = 14 Step 3 Use equation to solve for x. 2 2 2x + 4(2) = 14 x = 3 Solve for x. Substitute 2 for y. Check It Out! Example 2 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Step 4 Substitute for x and y in one of the original equations to solve for z. z = 1 Solve for z. 2x + 2y + 3z = 13 3 2(3) + 2(2) + 3z = 13 6 + 4 + 3z = 13 The solution to the system is (3, 2, 1). The points for first-place is 3, the points for second-place is 2, and 1 point for third-place. Check It Out! Example 2 Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Consistent means that the system of equations has at least one solution. Remember! The systems in Examples 1 and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Classify the system as consistent or inconsistent, and determine the number of solutions. Example 3: Classifying Systems with Infinite Many Solutions or No Solutions 2x – 6y + 4z = 2 –3x + 9y – 6z = –3 5x – 15y + 10z = 5 1 2 3

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Example 3 Continued 3(2x – 6y + 4z = 2) 2(–3x + 9y – 6z = –3) First, eliminate x. 1 2 6x – 18y + 12z = 6 –6x + 18y – 12z = –6 0 = 0 Multiply equation by 3 and equation by 2 and add. 21 The elimination method is convenient because the numbers you need to multiply the equations are small.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Example 3 Continued 5(2x – 6y + 4z = 2) –2(5x – 15y + 10z = 5) 1 3 10x – 30y + 20z = 10 –10x + 30y – 20z = –10 0 = 0 Multiply equation by 5 and equation by –2 and add. 3 1 Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent and has an infinite number of solutions.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3a Classify the system, and determine the number of solutions. 3x – y + 2z = 4 2x – y + 3z = 7 –9x + 3y – 6z = –12 1 2 3

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 3x – y + 2z = 4 –1(2x – y + 3z = 7) First, eliminate y. 1 3 3x – y + 2z = 4 –2x + y – 3z = –7 x – z = –3 Multiply equation by –1 and add to equation. 1 2 The elimination method is convenient because the numbers you need to multiply the equations by are small. Check It Out! Example 3a Continued 4

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 3(2x – y + 3z = 7) –9x + 3y – 6z = –12 2 3 6x – 3y + 9z = 21 –9x + 3y – 6z = –12 –3x + 3z = 9 Multiply equation by 3 and add to equation. 3 2 Now you have a 2-by-2 system. x – z = –3 –3x + 3z = 9 5 4 5 Check It Out! Example 3a Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 3(x – z = –3) –3x + 3z = 9 5 4 3x – 3z = –9 –3x + 3z = 9 0 = 0 Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent, and has an infinite number of solutions. Eliminate x. Check It Out! Example 3a Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3b Classify the system, and determine the number of solutions. 2x – y + 3z = 6 2x – 4y + 6z = 10 y – z = –2 1 2 3

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables y – z = –2 y = z – 2 3 Solve for y. Use the substitution method. Solve for y in equation 3. Check It Out! Example 3b Continued Substitute equation in for y in equation. 41 4 2x – y + 3z = 6 2x – (z – 2) + 3z = 6 2x – z + 2 + 3z = 6 2x + 2z = 4 5

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Substitute equation in for y in equation. 42 2x – 4y + 6z = 10 2x – 4(z – 2) + 6z = 10 2x – 4z + 8 + 6z = 10 2x + 2z = 2 6 Now you have a 2-by-2 system. 2x + 2z = 4 2x + 2z = 2 6 5 Check It Out! Example 3b Continued

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 2x + 2z = 4 –1(2x + 2z = 2) 6 5 Eliminate z. 0 2 Check It Out! Example 3b Continued Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables Lesson Quiz: Part I At the library book sale, each type of book is priced differently. The table shows the number of books Joy and her friends each bought, and the amount each person spent. Find the price of each type of book. paperback: $1; Hard- cover Paper- back Audio Books Total Spent Hal341$17 Ina251$15 Joy332$20 1. hardcover: $3; audio books: $4

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Holt McDougal Algebra 2 3-6 Solving Linear Systems in Three Variables 2. 3. 2x – y + 2z = 5 –3x +y – z = –1 x – y + 3z = 2 9x – 3y + 6z = 3 12x – 4y + 8z = 4 –6x + 2y – 4z = 5 inconsistent; none consistent; dependent; infinite Lesson Quiz: Part II Classify each system and determine the number of solutions.

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