Download HOMEWORK Chapter 1 Section 9

MTH 10905
Algebra
SOLVING LINEAR EQUATIONS
WITH A VARIABLE ON BOTH
SIDES OF THE EQUATION
CHAPTER 2 SECTION 5
Objectives
 Solving equations with the variable on both sides of the equal sign.
 Solve equations containing decimal numbers or fractions
 Indentify identities and contradictions.
Solving Linear Equations
 Solving Linear Equations with a variable on both sides of the equal sign.
 The general procedure is to “Isolate the variable”
 Steps
1.
2.
3.
4.
5.
6.
If the equation contains a fraction then multiply both sides by the LDC
Use the distributive property to remove any parentheses.
Combine any like terms that are on the same side of the equal sign.
Use the addition property giving you an equation in the form of ax = b
Use the multiplication property giving you x = b or 1x = b
a
a
Check your answer
Linear Equations - Solve
 There are several way that you can arrive at the correct answer.
The variable can be on the left side
 The variable can be on the right side
 You can choose which side you would like to have the
variable.

 I prefer to have the variable on whichever side makes it positive.
This avoids dividing by a negative.
 Once you have the variable by itself then the problem is just like
the ones that we have been solving in the previous sections.
Linear Equations - Solve
 Exp:
Check:
4 x  10  5 x  15
 10  5 x  4 x  15
 10  x  15
10  15  x
4 x  10  5 x  15
4(5)  10 ? 5(5)  15
20  10 ? 25  15
10  10
5 x
When we have isolated the variable on one side the equations
become like the ones that we solved in the previous section.
Linear Equations - Solve
 Exp:
Check:
4 x  2  6 x  28  8 x  4
 2 x  2  8 x  32
2  8 x  2 x  32
2  32  10 x
 30  10 x
30
x
10
3  x
4 x  2  6 x  28  8 x  4
(4)(3)  2  (6)(3)  28  (8)(3)  4
 12  2  (18)  28  24  4
 10  18  4  4
88
Linear Equations - Solve
 Exp:
3( p  2)  5 p  19
(3)( p )  (3)(2)  5 p  19
3 p  6  5 p  19
6  5 p  3 p  19
6  2 p  19
6  19  2 p
25  2 p
12
1
25
or 12.5 or
 p
2
2
Check:
 25 
 25 
3   2   (5)    19
2

2
 25 
 25 
(3)    (3)(2)  (5)    19
2
2
75
125
6
 19
2
2
75 12 125 38
 

2 2
2 2
87 87

or 43.5  43.5
2
2
1
43
or
2
Linear Equations - Solve
 Exp:
Check:
3( x  7)  2  4 x  4
3( x  7)  2  4 x  4
(3)( x)  (3)( 7)  2  4 x  4
3(23  7)  2  (4)(23)  4
3 x  21  2  4 x  4
(3)(23)  (3)(7)  2  (4)( 23)  4
3x  19  4 x  4
 69  21  2  92  4
 19  4  4 x  3 x
 90  2  88
 23  x
 88  88
Linear Equations - Solve
 Exp:
Check:
4  3 x  4 x  2( 4 x  5)
4  x  8 x  10
4  10  8 x  x
14  7 x
14
x
7
2x
4  3x  4 x  2(4 x  5)
4  (3)(2)  (4)(2)  2((4)(2)  5)
4  6  8  2(8  5)
 2  8  16  10
66
Linear Equations - Solve
 Exp:
Check:
8.25 x  6.50  4.25 x  1.22
8.25 x  4.25 x  1.22  6.50
4 x  7.72
7.72
x
4
x  1.93
8.25 x  6.50  4.25 x  1.22
8.25(1.93)  6.50  4.25(1.93)  1.22
 15.9225  6.50  8.2025  1.22
 9.4225  9.4225
Linear Equations - Solve
 Exp:
1
2
1
x 
x 
4
3
6
1
2
1
x 
x 
4
3
6
3
8
2
x 
x 
12
12
12

5
12
x 
2
12
1



2
12

 
x  
 12  
5 



1


x  
2
5
LCD  12
1 3 3
 
4 3 12
2 4 8
 
3 4 12
1 2 2
 
6 2 12
Linear Equations - Solve
 Exp:
x
 5  3( x  3)
5
x

5  5  3( x  3) 
5

 x
(5)   (5)(5)  5[3( x  3)]
5
x  25  15( x  3)
x  25  15 x  45
25  45  15 x  x
70  14 x
70 14 x

14 14
5 x
Linear Equations - Solve
 Exp:
x
 5  3( x  3)
5
1
x  5  3x  9
5
1

5  x  5  3x  9 
5

x  25  15 x  45
25  45  15 x  x
70  14 x
5 x
 5  1 
    55  53x   5 9
 1  5 
Linear Equations - Solve
3
1
 5 x  2   ( x  1)  2
5
2
 Exp:
15
6 1
1
x  x 2
5
5 2
2
1
1 6
 15

10 
x  x    2
2
2 5
 5

 15   5
(10 ) 
x    10
 5  
2
 1   5
  x    10
 2  
 1   2
     10
 2  
30 x  5 x  5  12  20
35 x  13
x
13
35
 6 
     10  2 
 5 
Linear Equations - Solve
 We have been studying Conditional equations they are
equations that have a single value for a solution. The are only
true under specific conditions. x = 3 or x = -5 or x = 2/3
 Identities are equations that are true for infinitely many values.
The solution process will end in one side of the equation
identical to the other. We write our answers as “all real
numbers” Exp: 0 = 0 or 4x + 6 = 4x + 6
 Contradictions are equations that have no solution. You will
obtain an obvious false statement. We write our answers as “no
Solution” Exp : 3 = 5 or
6=2
Linear Equations - Solve
 Exp:
7 x  6  3 x  4( x  1)  10
7 x  3 x  6  4 x  4  10
4x  6  4x  6
4x  4x  6  6
00
Identities: True, all real numbers
Linear Equations - Solve
 Exp:
7 x  4  3 x  5 x  10  9 x
7 x  3x  4  5 x  9 x  10
4 x  4  4 x  10
4 x  4 x  10  4
0  14
Contradiction: False, No solution
Linear Equations - Solve
 Conditional Equations have a single value for solution, only
one solution
 Identities have infinitely many values of x, many solutions
 Contradiction solution results in a false statement, have no
solution
HOMEWORK 2.5
 Page 138 - 139
#11, 17, 47, 63, 65, 75