Download Chapter 2 Section 4 - Lamar County School District

Chapter 2
Equations,
Inequalities and
Problem Solving
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
Bellwork:
1. 3x + 6 = 2x + 10 + x – 4 2. 6x + 2 – 2x = 4x + 1
3x + 6 = 3x + 6
0=0
a) Is 4 a solution of the eq.?
Yes
b) Is -3 a solution of the eq.?
2=1
0 = -1
a) Is -2 a solution of the eq.?
No
b) Is 6 a solution of the eq.?
Yes
c) Solve the equation to
determine its solution(s)?
True statement
All real numbers
No
c) Solve the equation to
determine its solution(s)?
False statement
No solution
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
2
2.4
Further Solving Linear
Equations
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
Objectives:
Apply
general strategy for
solving linear equations
Solve equations containing
fractions and decimals
Recognize identities and
equations with no solution
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
4
Solving Linear Equations
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
5
Example 1
Solve: 17 – x + 3 = 15 – (–6)
17  x  3  15  ( 6)
20  x  21
20  x  20  21  20
1x  1
1x 1

1 1
x  1
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
6
Example 2
3
Solve: 1  a  8
5
3
1 a  8
5
3
1 a 1  8 1
5
3
a7
5
5 3
5
 a  7
3 5
3
35
a
3
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
7
Example 3
Solve: 7(x – 3) = 9x – 7
7( x  3)  9 x  7
7 x  21  9 x  7
7 x  21  7 x  9 x  7  7 x
21  2 x  7
21  7  2 x  7  7
14  2 x
 14 2 x

2
2
7  x
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
8
Solving Linear Equations
Solve: 3( y  3)  2 y  6
5
5  3( y  3)
 52 y  6 
5
3 y  9  10 y  30
3 y  (3 y )  9  10 y  (3 y )  30
9  (30)  7 y  30  (30)
 21 7 y

7
7
3  y
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
9
Solving Linear Equations
Identity
5x – 5 = 2(x + 1) + 3x – 7
5x – 5 = 2x + 2 + 3x – 7
5x – 5 = 5x – 5
type of equation
which is always true
True Statement!
Both sides of the equation are identical. Since this
equation will be true for every x that is substituted into
the equation, the solution is “all real numbers.”
(-∞, ∞)
Infinitely many
solutions
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
10
Solving Linear Equations
Contradiction
3x – 7 = 3(x + 1)
3x – 7 = 3x + 3
type of equation
which is never true
3x + (–3x) – 7 = 3x + (–3x) + 3
–7 = 3
False Statement!
Since no value for the variable x
can be substituted into this
equation that will make this a true
statement, there is “no solution.”
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
11
Closure:
1.
What are the steps to solve
an equation?
2.
What is the difference
between an identity and a
contradiction?
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
12