Homework Log Wed 10/21 Lesson 3 – 5 Learning Objective: To solve systems with three variables Hw: Pg. 172 # odd Attention!! Pass up yesterday’s HW (worksheet – 3-2 word problems)
Lesson 3 – 5 Solving Systems with Three Variables Day 1 ALGEBRA II
Learning Objective 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 save (3)( ) (3) -12x + 3y – 15z = x – 3y + 3z = 31 -7x – 12z = -77 save
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 save (-5)( ) (-5) 2x + 4y – 5z = 18 25x – 15y + 5z = 85 27x – 11y = 103 save
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 save x – y – 2z = 4 -x + y – 3z = 11 -5z = 15 z = -3 nice!
(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 save (-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.