1. 2.2 Linear Systems in Three or More Variables

2. Solve using back-substitution.
x – 2y + 3z = 9 y + 3z = 5 z = 2
Sub. y = -1 and z = 2 into 1st equation.
Sub. z = 2 into 2nd equation.
y + 3(2) = 5
y + 6 = 5
y = -1
x – 2(-1) + 3(2) = 9
x = 9
x + 8 = 9
x = 1
Answer (x, y, z ) = (1, -1, 2)

3. Objective - To solve systems of linear equations in three variables.

4. Objective - To solve systems of linear equations in three variables.

5. Describe all the ways that three planes could
intersect in space.
Intersects at a Point
One Solution

6. Describe all the ways that three planes could
intersect in space.
Intersects at a Line
Infinitely Many Solutions

7. Describe all the ways that three planes could
intersect in space.
No Solution

8. Describe all the ways that three planes could
intersect in space.
No Solution

9. Solve.

10. Solve.

11. Identity Infinitely Many Solutions
Solve.
Identity Infinitely Many Solutions

12. In 1998, Cynthia Cooper of the WNBA Houston Comets basketball team was named Team Sportswoman of the Year. Cooper scored 680 points by hitting 413 of her 1-pt., 2-pt. and 3-point attempts. She made 40% of her pt. field goal attempts. How many 1-, 2- and 3-point baskets did Ms. Cooper make?
x = number of 1-pt. free throws y = number of 2-pt. field goals z = number of 3-pt. field goals x + y + z = 413 x + 2y + 3z = 680 z/160 = 0.4
-x - y - z = -413
x + 2y + 3z = 680
y + 2z = 267
y + 2(64) = 267
y = 139
x = 413
x = 210
z = 64

13. Find a quadratic function f(x) = ax2 + bx + c the graph of which passes through the points (-1, 3), (1, 1), and (2, 6).
Plug in each point for x and y. a(-1)2 + b(-1) + c = 3 a(1)2 + b(1) + c = 1 a(2)2 + b(2) + c = 6 Simplify a – b + c = 3 a + b + c = 1 4a + 2b + c = 6

14. Find a quadratic function f(x) = ax2 + bx + c the graph of which passes through the points (-1, 3), (1, 1), and (2, 6).
a – b + c = 3 a + b + c = 1 4a + 2b + c = 6
a – b + c = 3
a + b + c = 1
-2a - 2b - 2c = -2
4a + 2b + c = 6
2a + 2c = 4
2a – c = 4
2a – c = 4
2a + 2c = 4
-2a + c = -4
2a + 2c = 4
a – b + 0 = 3
a + b + 0 = 1
a – b = 3
a + b = 1
3c = 0
c = 0
2 + b + 0 = 1
b = -1
2a = 4
a = 2
f(x) = 2x2 – x