Linear Systems in Three or More Variables

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Solve using back-substitution.x – 2y + 3z = 9

y + 3z = 5

z = 2

Sub. z = 2 into 2nd equation.

y + 3(2) = 5y + 6 = 5y = -1

Sub. y = -1 and z = 2 into 1st equation.

x – 2(-1) + 3(2) = 9x + 2 + 6 = 9x + 8 = 9x = 1

Answer (x, y, z ) = (1, -1, 2)

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

Solve.

Describe all the ways that three planes could intersect in space.

Intersects at a Point

One Solution

Describe all the ways that three planes could intersect in space.

Intersects at a Line

Infinitely Many Solutions

Describe all the ways that three planes could intersect in space.

No Solution

Describe all the ways that three planes could intersect in space.

No Solution

Solve.

Solve.

Solve.

IdentityInfinitely Many Solutions

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 160 3-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

1                       y + 2z = 267

z = 64

y + 2(64) = 267 y = 139x + 139 + 64 = 413 x = 210

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 = 3a + b + c = 14a + 2b + c = 6

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 = 3a + b + c = 1

2a + 2c = 4

-2a - 2b - 2c = -2 4a + 2b + c = 6

2a – c = 4

2a – c = 42a + 2c = 4

-2a + c = -42a + 2c = 4

3c = 0c = 0

a – b + 0 = 3a + b + 0 = 1

a – b = 3 a + b = 1

2a = 4 a = 2

2 + b + 0 = 1 b = -1 f(x) = 2x2 – x