1. 3.4 See if you can figure this out: Can you replace the question marks with math symbols to make the following equation correct: (2 ? 3) ? (6 ? 2) ? (3 ? 1) = 5 Here are some possible answers: (2 x 3) + (6 / 2) - (3 + 1) = 5 (2 + 3) - (6 - 2) + (3 + 1) = 5 (2 - 3) + (6 - 2) + (3 - 1) = 5 (3 + 2) x (6 / 2) / (3 x 1) = 5

2. 3.4 The following is an example of a system of three linear equations in three variables: Equation 1:2x + y – z = 5 Equation 2:3x – 2y + z = 16 Equation 3:4x + 3y – 5z = 3 Reminder: The solution of a system of two equations was the point (x, y). The solution of a system of three linear equations is called an ordered triple and is written as: (x, y, z).

3. Solve the system Using elimination. 4x + 2y + 3z = 1 2x – 3y + 5z = – 14 6x – y + 4z = – 1 4x + 2y + 3z = 1 12x – 2y + 8z = – 2 16x + 11z = – 1 (Eliminate the variable with the smallest coefficient) Multiply by 2 No change 2x – 3y + 5z = – 14 – 18x + 3y – 12z = 3 – 16x – 7z = – 11 No change Multiply by -3 (Now choose the equation that you haven’t used yet and one other one – eliminate the same variable) (take the 2 new equations and eliminate another variable) 4z = –12 16x + 11z = – 1 16x + 11(-3) = – 1 16x + -33 = – 1 16x = 32z = – 3 x = 2 (Substitute z into one of the equations with only 2 variables) (Plug both variables back into any original equation) 6x – y + 4z = – 1 6(2) – y + 4(– 3) = – 1 12 – y – 12 = – 1 y = 1 Your solution is (2, 1, -3)

4. Solve the system using elimination. 3x + y – 2z = 10 6x – 2y + z = – 2 x + 4y + 3z = 7 The solution is (1, 3, – 2) 3x + y – 2z = 10 No change x 2 12x – 4y + 2z = – 4 15x – 3y = 6 x -3 No change -18x + 6y - 3z = 6 x + 4y + 3z = 7 -17x + 10y = 13 x 10 x 3 150x – 30y = 60 -51x + 30y = 39 99x = 99 15(1) – 3y = 6 15x – 3y = 6 15 – 3y = 6 – 3y = -9 x = 1 y = 3 6x – 2y + z = – 2 6(1) – 2(3) + z = – 2 6 – 6 + z = – 2 z = – 2

5. Solve a three-variable system with no solution x + y + z = 3 4x + 4y + 4z = 7 3x – y + 2z = 5 – 4x – 4y – 4z = – 12 4x + 4y + 4z = 7 0 = – 5 No change x -4 no solution false Solve a three-variable system with many solutions x + y + z = 4 x + y – z = 4 3x + 3y + z = 12 x + y + z = 4 x + y – z = 4 2x + 2y = 8 x + y – z = 4 3x + 3y + z = 12 4x + 4y = 16 No change -4x + -4y = -16 4x + 4y = 16 No change x -2 True infinitely many solutions 0 = 0

6. Solve the system. 2x – y – z = 15 4x + 5y + 2z = 10 -x – 4y + 3z = -20 Solution is (5, 0, -5).

7. Solve the system. 4x + 5y + 3z = 15 x – 3y + 2z = -6 -x + 2y – z = 3 HOMEWORK 3.4 p. 182 #5-30 (EOP) Solution: (2, 2, -1)