Warm Up Lesson Presentation Lesson Quiz. Holt Algebra 2 2

Transcription

1 3-6 Warm Up Lesson Presentation Lesson Quiz 2

2 Warm Up Solve each system of equations algebraically. x = 4y + 0 6x 5y = 9. (2, 2) 2. (, 3) 4x + 2y = 4 2x y = Classify each system and determine the number of solutions. 3x y = x 2y = 2 x = 3y 6x 2y = 4 inconsistent; none consistent, independent; one

3 Objectives Represent solutions to systems of equations in three dimensions graphically. Solve systems of equations in three dimensions algebraically.

4 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.

5 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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8 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.

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

10 Example Continued 2 5x 2y 3z = 7 3(2x 3y + z = 6) 5x 2y 3z = 7 6x 9y + 3z = 48 x y = 55 Multiply equation - 2 by 3, and add to equation. Use equations 3 and 2 to create a second equation in x and y x + 4y 2z = 7 2(2x 3y + z = 6) 3x + 4y 2z = 7 4x 6y + 2z = 32 7x 2y = 25 Multiply equation - 2 by 2, and add to equation 3. 5

11 Example Continued You now have a 2-by-2 system. x y = 55 7x 2y =

12 Example Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson (x y = 55) (7x 2y = 25) 22x + 22y = 0 77x 22y = x = 65 Multiply equation - 4 by 2, and equation - 5 by and add. x = 3 Solve for x.

13 Example Continued Step 3 Use one of the equations in your 2-by-2 system to solve for y. 4 x y = 55 ( 3) y = 55 y = 2 Substitute 3 for x. Solve for y.

14 Example Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2 2x 3y + z = 6 2( 3) 3(2) + z = 6 z = 4 The solution is ( 3, 2, 4). Substitute 3 for x and 2 for y. Solve for y.

15 Check It Out! Example Use elimination to solve the system of equations. x + y + 2z = 7 2x + 3y + z = 3x 4y + z = 4 Step Eliminate one variable. 2 3 In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is.

16 Check It Out! Example Continued 2 x + y + 2z = 7 x + y + 2z = 7 2(2x + 3y + z = ) 4x 6y 2z = 2 5x 5y = 5 4 Multiply equation - 2 by 2, and add to equation. Use equations and 3 to create a second equation in x and y. 3 x + y + 2z = 7 2( 3x 4y + z = 4) x + y + 2z = 7 6x + 8y 2z = 8 5x + 9y = Multiply equation - 3 by 2, and add to equation. 5

17 Check It Out! Example Continued You now have a 2-by-2 system. 5x 5y = 5 5x + 9y = 4 5

18 Check It Out! Example Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from Lesson x 5y = 5 5x + 9y = Add equation 5 to equation 4. 4y = 4 y = Solve for y.

19 Check It Out! Example Step 3 Use one of the equations in your 2-by-2 system to solve for x. 4 5x 5y = 5 5x 5() = 5 5x 5 = 5 5x = 0 x = 2 Substitute for y. Solve for x.

20 Check It Out! Example Step 4 Substitute for x and y in one of the original equations to solve for z. 2 2x +3y + z = 2( 2) +3() + z = z = z = 2 Substitute 2 for x and for y. Solve for z. The solution is ( 2,, 2).

21 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.

22 Example 2: Business Application 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. Orchestra Mezzanine Balcony Total Sales Fri $470 Sat $950 Sun $050

23 Example 2 Continued Step 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.

24 Step 2 Eliminate z. Example 2 Continued Multiply equation by 5 and equation 2 by 4 and add. 2 5(200x + 30y + 40z = 470) 4(250x + 60y + 50z = 950) 000x + 50y + 200z = x 240y 200z = 7800 y = 5 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.

25 Example 2 Continued Step 3 Use equation 3 to solve for x. 3 50x + 30y = 050 Substitute 5 for y. 50x + 30(5) = 050 Solve for x. x = 6

26 Example 2 Continued Step 4 Use equations or 2 to solve for z. 200x + 30y + 40z = 470 Substitute 6 for x and 5 for y. 200(6) + 30(5) + 40z = 470 z = 3 Solve for x. 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.

27 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 Winter Fair Chili Cook-off st Place 2nd Place 3rd Place Total Points Jada Maria Al

28 Check It Out! Example 2 Continued Step Let x represent first-place points, y represent second-place points, and z represent thirdplace points. Write a system of equations to represent the data in the table. 3x + y + 4z = 5 2x + 4y = 4 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 choic because eliminating z is straightforward.

29 Check It Out! Example 2 Continued Step 2 Eliminate z. Multiply equation by 3 and equation 3 by 4 and add. 3 3(3x + y + 4z = 5) 9x + 3y + 2z = 45 4(2x + 2y + 3z = 3) 8x 8y 2z = 52 x 5y = 7 4 Multiply equation 4 by 2 and add to equation (x 5y = 7) 2x + 4y = 4 2x + 0y = 4 2x + 4y = 4 y = 2 Solve for y.

30 Check It Out! Example 2 Continued Step 3 Use equation 2 to solve for x. 2 2x + 4y = 4 2x + 4(2) = 4 x = 3 Substitute 2 for y. Solve for x.

31 Check It Out! Example 2 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 3 2x + 2y + 3z = 3 2(3) + 2(2) + 3z = z = 3 z = Solve for z. The solution to the system is (3, 2, ). The points for first-place is 3, the points for second-place is 2, and point for third-place.

32 The systems in Examples and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions. Remember! Consistent means that the system of equations has at least one solution.

33 Example 3: Classifying Systems with Infinite Many Solutions or No Solutions Classify the system as consistent or inconsistent, and determine the number of solutions. 2x 6y + 4z = 2 3x + 9y 6z = 3 5x 5y + 0z = 5 2 3

34 Example 3 Continued The elimination method is convenient because the numbers you need to multiply the equations are small. First, eliminate x. 2 Multiply equation by 3 and equation 2 by 2 and add. 3(2x 6y + 4z = 2) 2( 3x + 9y 6z = 3) 6x 8y + 2z = 6 6x + 8y 2z = 6 0 = 0 ü

35 Example 3 Continued 3 Multiply equation by 5 and equation 3 by 2 and add. 5(2x 6y + 4z = 2) 0x 30y + 20z = 0 2(5x 5y + 0z = 5) 0x + 30y 20z = 0 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. ü

36 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 = 2 2 3

37 Check It Out! Example 3a Continued The elimination method is convenient because the numbers you need to multiply the equations by are small. First, eliminate y. Multiply equation 2 by and add to equation. 3 3x y + 2z = 4 (2x y + 3z = 7) 3x y + 2z = 4 2x + y 3z = 7 x z = 3 4

38 Check It Out! Example 3a Continued Multiply equation 2 by 3 and add to equation (2x y + 3z = 7) 9x + 3y 6z = 2 6x 3y + 9z = 2 9x + 3y 6z = 2 3x + 3z = 9 5 Now you have a 2-by-2 system. x z = 3 4 3x + 3z = 9 5

39 Eliminate x. Check It Out! Example 3a Continued 4 3(x z = 3) 3x 3z = 9 5 3x + 3z = 9 3x + 3z = 9 0 = 0 ü 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.

40 Check It Out! Example 3b Classify the system, and determine the number of solutions. 2x y + 3z = 6 2x 4y + 6z = 0 y z = 2 2 3

41 Check It Out! Example 3b Continued Use the substitution method. Solve for y in equation 3. 3 y z = 2 Solve for y. y = z 2 4 Substitute equation 4 in for y in equation. 2x y + 3z = 6 2x (z 2) + 3z = 6 2x z z = 6 2x + 2z = 4 5

42 Check It Out! Example 3b Continued Substitute equation 4 in for y in equation 2. 2x 4y + 6z = 0 2x 4(z 2) + 6z = 0 2x 4z z = 0 2x + 2z = 2 6 Now you have a 2-by-2 system. 2x + 2z = 4 5 2x + 2z = 2 6

43 Eliminate z. Check It Out! Example 3b Continued 5 6 2x + 2z = 4 (2x + 2z = 2) 0 2 û Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

44 . 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. Hardcover Paperback Audio Books Total Spent Hal 3 4 $7 Ina 2 5 $5 Joy $20 hardcover: $3; paperback: $; audio books: $4

45 Lesson Quiz: Part II Classify each system and determine the number of solutions. 2. 2x y + 2z = 5 3x +y z = x y + 3z = 2 inconsistent; none 3. 9x 3y + 6z = 3 2x 4y + 8z = 4 6x + 2y 4z = 5 consistent; dependent; infinite