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

Holt Algebra Solving Linear Systems in Three Variables

Holt Algebra Solving Linear Systems in Three Variables

Holt Algebra 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

Holt Algebra 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 = – x + 4y – 2z = 7 7x – 2y = –25 2(2x –3y + z = –16) 3x + 4y – 2z = 7 4x – 6y + 2z = – Multiply equation - by 3, and add to equation. 1 2 Multiply equation - by 2, and add to equation Use equations and to create a second equation in x and y. 3 2

Holt Algebra Solving Linear Systems in Three Variables 11x – 11y = –55 7x – 2y = –25 Step 2: You now have a 2-by-2 system. Solve for x and y. 4 5 Example 1 Continued

Holt Algebra Solving Linear Systems in Three Variables 2x – 3y + z = –16 2(–3) – 3(2) + z = – Step 3 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

Holt Algebra 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 = 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.

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

Holt Algebra 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

Holt Algebra Solving Linear Systems in Three Variables 4y = 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

Holt Algebra Solving Linear Systems in Three Variables –5x – 5(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

Holt Algebra Solving Linear Systems in Three Variables 2(–2) +3(1) + z = 1 2x +3y + z = 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 – z = 1

Holt Algebra 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 Fri $1470 Sat $1950 Sun150300$1050

Holt Algebra 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 = x + 60y + 50z = x + 30y = Friday’s sales. Saturday’s sales. Sunday’s 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.

Holt Algebra 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 x + 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.

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

Holt Algebra Solving Linear Systems in Three Variables 200x + 30y + 40z = Substitute 6 for x and 5 for y. 1 z = 3 Solve for x. Step 4 Use equations or to solve for z (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

Holt Algebra Solving Linear Systems in Three Variables Check It Out! Example 2 Jada’s 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 Maria Al Winter Fair Chili Cook-off

Holt Algebra 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 = Jada’s points. Maria’s points. Al’s points. A variable is “missing” in one equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

Holt Algebra 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 x + 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.

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

Holt Algebra 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 = (3) + 2(2) + 3z = z = 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

Holt Algebra 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.

Holt Algebra 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 =

Holt Algebra Solving Linear Systems in Three Variables Example 3 Continued 3(2x – 6y + 4z = 2) 2(–3x + 9y – 6z = –3) First, eliminate x x – 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.

Holt Algebra Solving Linear Systems in Three Variables Example 3 Continued 5(2x – 6y + 4z = 2) –2(5x – 15y + 10z = 5) x – 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.

Holt Algebra 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 = –

Holt Algebra Solving Linear Systems in Three Variables 3x – y + 2z = 4 –1(2x – y + 3z = 7) First, eliminate y x – 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

Holt Algebra Solving Linear Systems in Three Variables 3(2x – y + 3z = 7) –9x + 3y – 6z = – x – 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 = Check It Out! Example 3a Continued

Holt Algebra Solving Linear Systems in Three Variables 3(x – z = –3) –3x + 3z = x – 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

Holt Algebra 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 = –

Holt Algebra 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 x – y + 3z = 6 2x – (z – 2) + 3z = 6 2x – z z = 6 2x + 2z = 4 5

Holt Algebra 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 z = 10 2x + 2z = 2 6 Now you have a 2-by-2 system. 2x + 2z = 4 2x + 2z = Check It Out! Example 3b Continued

Holt Algebra 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.

Holt Algebra 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

Holt Algebra Solving Linear Systems in Three Variables x – 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.