Chapter Two 2.2

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Section 2.2
Linear Equations
Copyright © 2013 Pearson Education, Inc.
Page 100
If an equation is linear, writing it in the form ax + b = 0
should not require any properties or processes other
than the following:
• using the distributive property to clear parentheses,
• combining like terms,
• applying the addition property of equality.
Example
Page 100
Determine whether the equation is linear. If the equation
is linear, give values for a and b.
5
3
a. 9x + 7 = 0
b. 5x + 9 = 0
c. 7   0
x
Solution
a. The equation is linear because it is in the form
ax + b = 0 with a = 9 and b = 7.
b. This equation is not linear because it cannot be written
in the form ax + b = 0. The variable has an exponent
other than 1.
c. This equation is not linear because it cannot be written
in the form ax + b = 0. The variable appears in the
denominator of a fraction.
Example
Page 101
Complete the table for the given values of x. Then solve
the equation 4x – 6 = −2.
x
−3
4x − 6
−18
−2
−1
0
1
2
3
Solution
x
4x − 6
−3
−2
−1
0
1
2
3
−18 −14
−10
−6
−2
2
6
From the table 4x – 6 = −2 when x = 1. Thus the
solution to 4x – 6 = −2 is 1.
Solving a Linear Equation Symbolically
Page 102
STEP 1: Use the distributive property to clear any parentheses on
each side of the equation. Combine any like terms on
each side.
STEP 2: Use the addition property of equality to get all of the
terms containing the variable on one side of the
equation and all other terms on the other side of the
equation. Combine any like terms on each side.
STEP 3: Use the multiplication property of equality to isolate the
variable by multiplying each side of the equation by the
reciprocal of the number in front of the variable (or divide
each side by that number).
STEP 4: Check the solution by substituting it in the given
equation.
Slide 5
Example
Page 103
Solve each linear equation.
a. 12x − 15 = 0
b. 3x + 19 = 5x + 5
Solution
a. 12x – 15 = 0
b. 3x + 19 = 5x + 5
12x − 15 + 15 = 0 + 15
3x − 3x + 19 = 5x − 3x + 5
19 = 2x + 5
12x = 15
19 − 5 = 2x + 5 − 5
12 x 15

14 = 2x
12
12
5
x
4
14 2x

2
2
7x
Example
Page 104
Solve the linear equation. Check your solution.
x + 5 (x – 1) = 11
Solution
x + 5 (x – 1) = 11
x + 5x – 5 = 11
6x − 5 = 11
6x − 5 + 5 = 11 + 5
6x = 16
6 x 16

6
6
16 8
x

6 3
Example (cont)
Page 104
Check: x + 5 (x – 1) = 11
8
8 
 5   1  11
3
3 
8
8 3
 5     11
3
3 3
8
5
 5    11
3
3
8 25

 11
3 3
33
 11
3
11  11
The answer checks, the
8
solution is .
3
Example
Page 105
1
1
Solve the linear equation. x  x  5
2
6
Solution
1
1
x x 5
2
6
1 
1
6 x  x   56
6 
2
3x  x  30
2x  30
x  15
The solution is 15.
Example
Page 106
Solve the linear equation. 5.3  0.8x  7
Solution
5.3  0.8x  7
10  5.3  0.8x   7 10
53  8x  70
53   53  8x  70  53
8x  17
8 x 17

8
8
17
x
8
17
The solution is  .
8
Page 107
Equations with No Solutions or Infinitely Many
Solutions
An equation that is always true is called an identity and
an equation that is always false is called a contradiction.
Slide 11
Example
Page 107
Determine whether the equation has no solutions, one
solution, or infinitely many solutions.
a. 10 – 8x = 2(5 – 4x)
b. 7x = 9x + 2(12 – x)
c. 6x = 4(x + 5)
Solution
10 – 8x = 2(5 – 4x)
10 – 8x = 10 – 8x
– 8x = – 8x
0=0
Because the equation 0 = 0 is
always true, it is an identity
and there are infinitely many
solutions.
Example (cont)
b. 7x = 9x + 2(12 – x)
Page 107
c. 6x = 4(x + 5)
7x = 9x + 24 – 2x
6x = 4x + 20
7x = 7x + 24
2x = 20
0 = 24
Because the equation
0 = 24 is always false, it is
a contradiction and there
are no solutions.
x = 10
Thus there is one
solution.
Page 107
Slide 14
End of chapter problems
•
do 42, 44, 46, 56 on page 109
•
do 69, 70 on page 109
DONE
Objectives
•
Basic Concepts
•
Solving Linear Equations
•
Applying the Distributive Property
•
Clearing Fractions and Decimals
•
Equations with No Solutions or Infinitely Many Solutions
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