Download 9.3 Equations as Relations

9.3 Equations as
Relations
CORD Math
Mrs. Spitz
Fall 2006
Objectives:
• Solve linear equations for a specific
variable, and
• Solve linear equations for a given domain
• What’s linear?
Assignment
• Pgs. 367 -368 #4 – 41 all
Definition of the solution of
an equation in two variables
• If a true statement results when the
numbers in an ordered pair are substituted
into an equation in two variables, then the
ordered pair is a solution of the equation
More
• Since the solutions of an equation in two
variables are ordered pairs, such an
equation describes a relation. The set of
values of x is the domain of the relation.
The set of corresponding values of y is the
range of the relation.
Ex. 1: Solve y = 2x +3 if the domain is
{-5, -3, -1, 0, 1, 3, 5, 7, 9}
x
2x + 3
y
(x, y)
-5
2(-5) + 3 = -10 + 3 = -7
-7
(-5, -7)
-3
2(-3) +3 = -6 + 3 = -3
-3
(-3, -3)
-1
2(-1) + 3 = -2 + 3 = 1
1
(-1, 1)
0
2(0) + 3 = 0 + 3 = 3
3
(0, 3)
1
2(1) + 3 = 2 + 3 = 5
5
(1, 5)
3
2(3) + 3 = 6 + 3 = 9
9
(3, 9)
5
2(5) + 3 = 10 + 3 = 13
13
(5, 13)
7
2(7) + 3 = 14 + 3 = 17
17
(7, 17)
9
2(9) + 3 = 18 + 3 = 21
21
(9, 21)
The solution set is (x, y) column
Ex. 2: Solve 3y + 6x = 12 if the domain is
{-4, -3, -2, 2, 3, 4}
3y + 6x = 12
3y = -6x + 12
First solve for y:
y = -2x + 4
x
-2x + 4
y
(x, y)
-4
-2(-4) + 4 = 8 + 4 = 12
12
(-4, 12)
-3
-2(-3) +4 = 6 + 4 = 10
10
(-3, 10)
-2
-2(-2) + 4 = 4 + 4 = 8
8
(-2, 8)
2
-2(2) + 4 = -4 + 4 = 0
0
(2, 0)
3
-2(3) + 4 = -6 + 4 = -2
-2
(3, -2)
4
-2(4) + 4 = -8 + 4 = -4
-4
(4, -4)
The solution set is (x, y) column
Homework problems
#4,5 are table of values to complete
#6-8/13-15 are multiple choice.
#9-12/16-23 are solve for the variable
indicated
#24-41 are given that domain is, then solve
each equation for the range.
You need a ruler to make tables or be
careful making them.