Download Objective - to solve systems of linear equations in three variables.

Linear Systems in Three or More
Variables
(teacherweb.com)
Solve using back-substitution.
x – 2y + 3z = 9
y + 3z = 5
z=2
Sub. z = 2 into
2nd
y + 3(2) = 5
y+6=5
y = -1
equation.
Sub. y = -1 and z = 2 into 1st
equation.
x – 2(-1) + 3(2) = 9
x+2+6=9
x+8=9
x=1
Answer (x, y, z ) = (1, -1, 2)
Objective - To solve systems of linear equations
in three variables.
Solve.
2x + y + z = 3
3x + 2y - z = 2
2x + y + z = 3
+
3x + 2y - z = 2
-x + 3y + 2z = 1
+ 2y - z = 2)
2 (3x
6x + 4y - 2z = 4
+-x + 3y + 2z = 1
-x + 3y + 2z = 1
-
5x + 3 y = 5
5x + 7 y = 5
-4 y = 0
5x + 3 y = 5
y=0
5 x + 7 (0 ) = 5
x =1
5x + 7 y = 5
2(1) + 0 + z = 3
z = 1 (1,0,1)
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.
- 2y + z = 7)
-2 (x -2x + 4y - 2z = -14
+
2x + y + 2z = 9
2x + y + 2z = 9
5 y = -5
-2x + 3y + 4z = -1
y = -1
2x + y + 2z = 9
+
-2x + 3y + 4z = -1
4 (-1) + 6 z = 8
4 y + 6z = 8
z=2
x - 2(-1) + 2 = 7
x=3
(3,-1,2)
Solve.
+ 2y + 2z = 10)
3(2x
6x + 6y + 6z = 15
6x + 6y + 6z = 30
-
6x + 6y + 6z = 15
0 = 15
x - 3y - z = 8
No Solution
Solve.
x+ y+z= 4
x+ y-z= 4
3x + 3y + z = 12
x+ y-z = 4
+
3x + 3y + z = 12
4 x + 4 y = 16
x+ y+z= 4
+ x + y -z = 4
2x + 2 y = 8
x+ y =4
-x + y = 4
Identity
Infinitely Many Solutions
0=0
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 1pt., 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
1
x + y + z = 413
x + 2y + 3z = 680
z/160 = 0.4

z = 64
-x - y - z = -413
x + 2y + 3z = 680
y + 2z = 267
y + 2(64) = 267
y = 139
x + 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=3
a+b+c=1
4a + 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).
-2a - 2b - 2c = -2
4a + 2b + c = 6
2a – c = 4
2a – c = 4
2a + 2c = 4
2+b+0=1
b = -1
a–b+c=3
a+b+c=1
4a + 2b + c = 6
-2a + c = -4
2a + 2c = 4
3c = 0
c=0
f(x) = 2x2 – x
a–b+c=3
a+b+c=1
2a + 2c = 4
a–b+0=3
a+b+0=1
a–b=3
a+b=1
2a = 4
a=2