Use elimination to solve the system of equations.

1. 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 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.

2. 1 2 3 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. Ex 1: Use elimination to solve the system of equations. 5x – 2y – 3z = –7 2x – 3y + z = –16 3x + 4y – 2z = 7 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.

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

4. 4 5 4 4 5 Multiply equation - by –2, and equation - by 11 and add. 4 5 11x – 11y = –55 You now have a 2-by-2 system. 7x – 2y = –25 Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. –2(11x – 11y = –55) –22x + 22y = 110 11(7x – 2y = –25) 77x – 22y = –275 1 55x = –165 x = –3 1 Step 3 Use one of the equations in your 2-by-2 system to solve for y. 11x – 11y = –55 11(–3) – 11y = –55 y = 2

5. 2 Step 4 Substitute for x and y in one of the original equations to solve for z. 2x – 3y + z = –16 2(–3) – 3(2) + z = –16 z = –4 1 The solution is (–3, 2, –4). 1 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.

6. 1 2 3 Ex 2: 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. 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. 250x + 60y + 50z = 1950 Saturday’s sales. 150x + 30y = 1050 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.

7. Multiply equation by 5 and equation by –4 and add. 1 2 1 1 2 3 2 3 1 Step 3 Use equation to solve for x. Step 4 Use equations or to solve for z. Step 2 Eliminate z. 1000x + 150y + 200z = 7350 5(200x + 30y + 40z = 1470) –1000x – 240y – 200z= –7800 –4(250x + 60y + 50z = 1950) y = 5 1 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y. 150x + 30y= 1050 200x + 30y + 40z = 1470 150x + 30(5)= 1050 200(6) + 30(5)+ 40z = 1470 x = 6 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.

8. 1 2 3 1 2 1 2 Multiply equation by 3 and equation by 2 and add. Ex 3: Classify the system as consistent or inconsistent, and determine the number of solutions. 2x – 6y + 4z = 2 –3x + 9y – 6z = –3 5x – 15y + 10z = 5 The elimination method is convenient because the numbers you need to multiply the equations are small. First, eliminate x. 6x – 18y + 12z = 6 3(2x – 6y + 4z = 2) 2(–3x+ 9y – 6z = –3) –6x + 18y – 12z = –6 0 = 0 

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