Linear systems with 3 unknows

Solving Linear Systems in Three Variables Warm Up Lesson Presentation Lesson Quiz

Warm Up Solve each system of equations algebraically. Classify each system and determine the number of solutions. 1. 2. (2, –2) (– 1, – 3) 3. 4. inconsistent; none consistent, independent; one x = 4 y + 10 4 x + 2 y = 4 6 x – 5 y = 9 2 x – y =1 3 x – y = 8 6 x – 2 y = 2 x = 3 y – 1 6 x – 12 y = –4

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

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.

Recall from previous lessons 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.

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 previous lessons.

Use elimination to solve the system of equations. Example 1: Solving a Linear System in Three Variables Step 1 Eliminate one variable. 5 x – 2 y – 3 z = –7 2 x – 3 y + z = –16 3 x + 4 y – 2 z = 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

Example 1 Continued 5 x – 2 y – 3 z = –7 11 x – 11 y = –55 3( 2 x –3 y + z = –16 ) 5 x – 2 y – 3 z = –7 6 x – 9 y + 3 z = –48 1 3 x + 4 y – 2 z = 7 7 x – 2 y = –25 2( 2 x –3 y + z = –16 ) 3 x + 4 y – 2 z = 7 4 x – 6 y + 2 z = –32 1 2 4 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

11 x – 11 y = –55 7 x – 2 y = –25 You now have a 2-by-2 system. Example 1 Continued 4 5

– 2( 11 x – 11 y = –55 ) 55 x = –165 11( 7 x – 2 y = –25 ) – 22 x + 22 y = 110 77 x – 22 y = –275 1 1 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from previous lessons. x = –3 Solve for x. Example 1 Continued 4 5 Multiply equation - by – 2, and equation - by 11 and add. 4 5

11 x – 11 y = –55 11 (–3) – 11 y = –55 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 4

2 x – 3 y + z = –16 2 (–3) – 3 (2) + z = –16 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 2

Use elimination to solve the system of equations. Step 1 Eliminate one variable. – x + y + 2 z = 7 2 x + 3 y + z = 1 – 3 x – 4 y + z = 4 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. 1 2 3

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

You now have a 2-by-2 system. Check It Out! Example 1 Continued – 5 x – 5 y = 5 5 x + 9 y = –1 4 5

4 y = 4 1 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from previous lessons. Solve for y. Check It Out! Example 1 Continued – 5 x – 5 y = 5 5 x + 9 y = –1 y = 1 4 5 Add equation to equation . 4 5

– 5 x – 5 (1) = 5 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 – 5 x – 5 y = 5 – 5 x – 5 = 5 – 5 x = 10 4

2 (–2) +3 (1) + z = 1 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 2 x +3 y + z = 1 2

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.

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 Orchestra Mezzanine Balcony Total Sales Fri 200 30 40 $1470 Sat 250 60 50 $1950 Sun 150 30 0 $1050

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. 200 x + 30 y + 40 z = 1470 250 x + 60 y + 50 z = 1950 150 x + 30 y = 1050 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. 1 2 3

5( 200 x + 30 y + 40 z = 1470 ) – 4( 250 x + 60 y + 50 z = 1950 ) Step 2 Eliminate z. 1000 x + 150 y + 200 z = 7350 – 1000 x – 24 0 y – 200 z = –7800 y = 5 Example 2 Continued By eliminating z , due to the coefficients of x , you also eliminated x providing a solution for y . 1 Multiply equation by 5 and equation by –4 and add. 1 2 2

150 x + 30 y = 1050 150 x + 30 (5) = 1050 Substitute 5 for y. x = 6 Solve for x. Example 2 Continued 3 Step 3 Use equation to solve for x. 3

200 x + 30 y + 40 z = 1470 Substitute 6 for x and 5 for y. 1 z = 3 Solve for x. 200 (6) + 30 (5) + 40 z = 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 1 Step 4 Use equations or to solve for z. 2 1

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 3 1 4 15 Maria 2 4 0 14 Al 2 2 3 13 Winter Fair Chili Cook-off

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. 3 x + y + 4 z = 15 2 x + 4 y = 14 2 x + 2 y + 3 z = 13 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. 1 2 3

3( 3 x + y + 4 z = 15 ) – 4( 2 x + 2 y + 3 z = 13 ) Step 2 Eliminate z. 9 x + 3 y + 12 z = 45 – 8 x – 8 y – 12 z = –52 x – 5 y = –7 Check It Out! Example 2 Continued – 2 ( x – 5 y = –7 ) 2 x + 4 y = 14 – 2 x + 10 y = 14 2 x + 4 y = 14 y = 2 Solve for y. 1 Multiply equation by 3 and equation by –4 and add. 3 1 3 4 2 4 Multiply equation by –2 and add to equation . 2 4

2 x + 4 y = 14 2 x + 4 (2) = 14 x = 3 Solve for x. Substitute 2 for y. Check It Out! Example 2 Continued Step 3 Use equation to solve for x. 2 2

Step 4 Substitute for x and y in one of the original equations to solve for z . z = 1 Solve for z. 2 (3) + 2 (2) + 3 z = 13 6 + 4 + 3 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 2 x + 2 y + 3 z = 13 3

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. Consistent means that the system of equations has at least one solution. Remember!

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

Example 3 Continued 3( 2 x – 6 y + 4 z = 2 ) 2( –3 x + 9 y – 6 z = –3 ) First, eliminate x . 6 x – 18 y + 12 z = 6 – 6 x + 18 y – 12 z = –6 0 = 0 The elimination method is convenient because the numbers you need to multiply the equations are small.  1 2 Multiply equation by 3 and equation by 2 and add. 2 1

Example 3 Continued 5( 2 x – 6 y + 4 z = 2 ) – 2( 5 x – 15 y + 10 z = 5 ) 10 x – 30 y + 20 z = 10 – 10 x + 30 y – 20 z = –10 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.  1 3 Multiply equation by 5 and equation by –2 and add. 3 1

Check It Out! Example 3a Classify the system, and determine the number of solutions. 3 x – y + 2 z = 4 2 x – y + 3 z = 7 – 9 x + 3 y – 6 z = –12 1 2 3

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

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

3( x – z = –3 ) – 3 x + 3 z = 9 3 x – 3 z = –9 – 3 x + 3 z = 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 5 4

Check It Out! Example 3b Classify the system, and determine the number of solutions. 2 x – y + 3 z = 6 2 x – 4 y + 6 z = 10 y – z = –2 1 2 3

y – z = –2 y = z – 2 Solve for y. Use the substitution method. Solve for y in equation 3. Check It Out! Example 3b Continued 2 x – y + 3 z = 6 2 x – ( z – 2) + 3 z = 6 2 x – z + 2 + 3 z = 6 2 x + 2 z = 4 3 Substitute equation in for y in equation . 4 1 4 5

2 x – 4 y + 6 z = 10 2 x – 4( z – 2) + 6 z = 10 2 x – 4 z + 8 + 6 z = 10 2 x + 2 z = 2 Now you have a 2-by-2 system. 2 x + 2 z = 4 2 x + 2 z = 2 Check It Out! Example 3b Continued Substitute equation in for y in equation . 4 2 6 6 5

2 x + 2 z = 4 – 1( 2 x + 2 z = 2 ) 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. 6 5

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; 1. hardcover: $3; audio books: $4 Hard-cover Paper- back Audio Books Total Spent Hal 3 4 1 $17 Ina 2 5 1 $15 Joy 3 3 2 $20

2. 3. 2 x – y + 2 z = 5 – 3 x + y – z = –1 x – y + 3 z = 2 9 x – 3 y + 6 z = 3 12 x – 4 y + 8 z = 4 – 6 x + 2 y – 4 z = 5 inconsistent; none consistent; dependent; infinite Lesson Quiz: Part II Classify each system and determine the number of solutions.

Linear systems with 3 unknows

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Linear systems with 3 unknows