Lesson 3–5,3–6 Objectives Be able to solve “3-by-3” systems of equations algebraically

A Global Positioning System (GPS) gives locations using the three coordinates of latitude, longitude, and elevation. You can represent any location in three-dimensional space using a three-dimensional coordinate system, sometimes called coordinate space.

Each point in coordinate space can be represented by an ordered triple of the form (x, y, z). The system is similar to the coordinate plane but has an additional coordinate based on the z-axis. Notice that the axes form three planes that intersect at the origin.

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.

The graph of a linear equation in three variables is a plane 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 Lesson 3-2.

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

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

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

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

Step 3 Use one of the equations in your 2-by-2 system to solve for y. Example 1 Continued Step 3 Use one of the equations in your 2-by-2 system to solve for y. 11x – 11y = –55 4 1 11(–3) – 11y = –55 Substitute –3 for x. 1 y = 2 Solve for y.

Substitute –3 for x and 2 for y. 2(–3) – 3(2) + z = –16 Example 1 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2x – 3y + z = –16 2 Substitute –3 for x and 2 for y. 2(–3) – 3(2) + z = –16 1 z = –4 Solve for y. 1 The solution is (–3, 2, –4).

Example 1: Solving a Linear System in Three Variables Use elimination to solve the system of equations. –x + y + 2z = 7 1 2x + 3y + z = 1 2 –3x – 4y + z = 4 3 Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.

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

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

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

Step 3 Use one of the equations in your 2-by-2 system to solve for x. Check It Out! Example 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. –5x – 5y = 5 4 Substitute 1 for y. –5x – 5(1) = 5 1 Solve for x. –5x – 5 = 5 –5x = 10 1 x = –2

Substitute –2 for x and 1 for y. 2x +3y + z = 1 Check It Out! Example 1 Step 4 Substitute for x and y in one of the original equations to solve for z. Substitute –2 for x and 1 for y. 2x +3y + z = 1 2 2(–2) +3(1) + z = 1 –4 + 3 + z = 1 Solve for z. 1 z = 2 1 The solution is (–2, 1, 2).

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 200 30 40 $1470 Sat 250 60 50 $1950 Sun 150 $1050

Write a system of equations to represent the data in the table. 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 Friday’s sales. 1 250x + 60y + 50z = 1950 Saturday’s sales. 2 150x + 30y = 1050 Sunday’s sales. 3 A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

Multiply equation by 5 and equation by –4 and add. Example 2 Continued Step 2 Eliminate z. Multiply equation by 5 and equation by –4 and add. 1 2 5(200x + 30y + 40z = 1470) 1000x + 150y + 200z = 7350 1 –4(250x + 60y + 50z = 1950) –1000x – 240y – 200z = –7800 2 y = 5 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.

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

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

Lesson Assignment Read Lessons* ----------------------------------------------------------- Lesson 3-6 “Practice B” Worksheet #2, 4, 6, 7, 8, 9