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Wed 10/21 Lesson 3 – 5 Learning Objective: To solve systems with three variables Hw: Pg. 171 # 21 – 29 odd

Algebra II

 To solve systems with three variables

 Use two eq’ns to get rid of one variable  Use two different eq’ns to get rid of SAME variable  Line up eq’ns resulting from first two steps & solve  Plug in to find remaining variables

(2)( ) (2) 3x – 2y + 4z = 35 -8x + 2y – 10z = x – 6z = -37 (3)( ) (3) -12x + 3y – 15z = x – 3y + 3z = 31 -7x – 12z = -77

10x + 12z =74 -7x – 12z = -77 3x = -3 x = -1 (-2)( ) (-2) (-1, -5, 7) -5(-1) - 6z = – 6z = z = - 42 z = 7 3(-1) – 2y + 4(7) = – 2y + 28 = 35 -2y + 25 = 35 -2y = 10 y = -5

(2)( ) (2) -3x + 5y + 2z = x + 6y – 2z = x + 11y = -61 (-5)( ) (-5) 2x + 4y – 5z = 18 25x – 15y + 5z = 85 27x – 11y = 103

14x = 42 x = 3 (3, -2, -4) -13(3) + 11y = y = y = -22 y = -2 2(3) + 4(-2) – 5z = 18 6 – 8 – 5z = – 5z = 18 -5z = 20 z = -4

x – y – 2z = 4 -x + 2y + z = 1 y - z = 5 x – y – 2z = 4 -x + y – 3z = 11 -5z = 15 z = -3

(0, 2, -3) x – 2 – 2(-3) = 4 x – = 4 x + 4 = 4 x = 0 3. y - (-3) = 5 y + 3 = 5 y = 2

Since one eq’n is already missing a “x”, eliminate the x in the other two eq’ns Get variables on  same side! x – 2y – 3z = 0

x – y + 2z = -7 -x + 2y + 3z = 0 y + 5z = -7 (-1)( ) (-1)

y + z = 1 -y – 5z = 7 -4z = 8 z = -2 (0, 3, -2) y + (-2) = 1 y = 3 x – (3) + 2(-2) = -7 x – 3 – 4 = -7 x – 7 = -7 x = 0 (-1)( ) (-1)

One variable is already given! Start plugging! 4 + 5y = 9 5y = 5 y = 1 2(4) + 3(1) – 2z = – 2z = – 2z = -1 -2z = -12 z = 6 (4, 1, 6)

 A classmate says that the system consisting of x = 0, y = 0, and z = 0 has no solution. Explain the student’s error.