Download Solving a Linear Equation

Section 2.3
Solving Linear
Equations
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Objective 1
Solve linear equations.
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Solving a Linear Equation
1. Simplify the algebraic expression on
each side.
2. Collect all the variable terms on one side
and all the constant terms on the other side.
3. Isolate the variable and solve.
4. Check the proposed solution in the
original equation.
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Example
Solve and check: 7 x  3  x  8 
7x  3  x  8
Use the Distributive
7 x  3 x  24
Property.
7 x  3 x  3 x  24  3 x Subtract 3x from
both sides.
Simplify.
4 x  24
Solve.
x6
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Example (cont)
7  6   3  6  8  Check the solution in the
original equation.
42  3 14 
42  42
The true statement 42  42
verifies 6 is the solution.
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Objective 1: Example
1a.
Solve: 7 x  25  3 x  16  2 x  3
7 x  25  3 x  16  2 x  3
4 x  25  13  2 x
Simplify the algebraic
expression on each side.
4 x  25  2 x  13  2 x  2 x Collect variable terms on
one side and constant
terms on the other side.
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Objective 1: Example (cont)
2 x  25  13
2 x  25  25  13  25
2 x  12
2 x 12

Isolate the variable and solve.
2
2
x6
The solution set is 6.
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Objective 1: Example
1b.
Solve: 4(2 x  1)  29  3(2 x  5)
4(2 x  1)  29  3(2 x  5)
8 x  4  29  6 x  15
Simplify the algebraic
expression on each side.
8 x  25  6 x  15
8 x  6 x  25  6 x  6 x  15 Collect variable terms on
one side and constant
terms on the other side.
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Objective 1: Example (cont)
2 x  25  15
2 x  25  25  15  25
2 x  10
2 x 10

Isolate the variable and solve.
2
2
x 5
The solution set is 5.
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Objective 2
Solve linear equations containing fractions.
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Example
2x  1 x  2 x
Solve & Check:


3
5
2
1. Simplify the algebraic expressions on each
side.
2x  1 x  2 x


3
5
2
 2x  1 x  2 
 x  Multiply both sides
30 

 30  

5 
 3
 2  By LCD: 30
30  2 x  1  30  x  2  30  x  Distributive






1  3  1  5  1  2  Property
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Example (cont)
30  2 x  1  30  x  2  30  x 








1  3  1  5  1 2
10  2 x  1  6  x  2  15  x 
 






1  1  1 1  1  1
10  2 x  1  6  x  2   15 x
20 x  10  6 x  12  15 x
14 x  2  15 x
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Cancel
Multiply
Distribute
Combine like
terms
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Example (cont)
2. Collect variable terms on one side and
constant terms on other side.
14 x  14 x  2  15 x  14 x Subtract 14x from
both sides
2x
Simplify
3. Isolate the variable and solve. Already done.
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Example (cont)
4. Check the proposed solution in the original
equation.
2x  1 x  2

3
5
2  2  1  2  2

3
5
4 1 2  2

3
5
3 0

3 5

?

?

?

x
2
2
2
2
2
2
2
Original equation
Replace x with 2
Simplify
Simplify
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Example (cont)
?
1 0  1
1 1
Simplify
Simplify
Since the proposed x value of 2 makes the
statement true, 2 is indeed the solution of the
original equation.
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Objective 2: Example
2.
Solve:
x
4
x
12 
4
x
12 
4
3x
x 2x 5


4
3 6
2x 5


3 6
 2x 5 
 12 
 
 3 6
2x
5
 12 
 12 
3
6
 8 x  10
Begin by multiplying
both sides of the
equation by 12, the
least common
denominator
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Objective 2: Example (cont)
3x  8x
5 x
5 x
5
x
 8 x  8 x  10
 10
10

5
 2
The solution set is 2.
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Objective 3
Solve linear equations containing decimals.
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Clearing an Equation of Decimals
Multiply every term on both sides of the
equation by a power of 10. The exponent on
10 will equal the greatest number of decimal
places in the equation.
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Objective 3: Example
3.
Solve: 0.48 x  3  0.2  x  6 
First apply the distributive property to remove
the parentheses, and then multiply both
sides by 100 to clear the decimals.
0.48 x  3  0.2  x  6 
0.48 x  3  0.2 x  1.2
100  0.48 x  3   100  0.2 x  1.2 
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Objective 3: Example (cont)
48 x  300  20 x  120
48 x  300  300  20 x  120  300
48 x  20 x  420
48 x  20 x  20 x  20 x  420
28 x  420
28 x
480

28
28
x  15
The solution set is 15.
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Objective 4
Identify solutions with no solution or infinitely
many solutions.
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Linear Equations having No Solutions
Some equations are not true for any real
number. Such equations are call
inconsistent equations or contradictions.
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Example
Solve:
2x  2  x  3 
2x  2x  6
Distributive Property
2 x  2 x  2 x  2 x  6 Subtract 2 x from both
sides
06
Simplify. False, 0  6
Inconsistent, no solution.
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Categorizing an Equation
Type of Equations
Identity
Conditional
Inconsistent
(contradiction)
Definitions
An equation that is true
for all real numbers
An equation that is not
an identity but is true for
at least one real number
An equation that is not
true for any real number
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Example
Solve and determine whether the equation is an
identity, a conditional equation or an inconsistent
equation: 5  4 x  9 x  5
5  4x
5  5  4x
4x
4x  4x
 9x  5
 9 x  5  5 Subtract 5 from both sides
 9x
Simplify
 9x  4x
Subtract 4 x from both sides
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Example (cont)
0  5x
0 5x

5
5
0x
Simplify
Divide both sides by 5
Simplify
The original equation is only true when
x  0. Therefore, it is a conditional equation.
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Example
Solve and determine whether the equation is an
identity, a conditional equation or an inconsistent
equation: 5   2 x  4   4  x  1  2 x
5  (2 x  4)  4( x  1)  2 x
5  2 x  4  4 x  4  2 x Distribute the  1 and the 4.
9  2x  4  2x
Simplify.
94
Add 2 x to both sides.
Since after simplification we see a contradiction,
we know that the original equation is
inconsistent and can never be true for any x.
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Example
Solve and determine whether the equation is an
identity, a conditional equation or an inconsistent
equation: 3  2 x  3  x  1  x
3  2 x  3  x  1  x
3  2x  3 x  3  x
Distribute the 3.
3  2x  2x  3
Simplify.
Since after simplification, we can see that the left hand
side (LHS) is equal to the RHS of the equation, this is
an identity and is always true for all x.
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Objective 4: Example
4a.
Solve: 3 x  7  3  x  1
3 x  7  3  x  1
3x  7  3x  3
3x  3x  7  3x  3x  3
73
The original equation is equivalent to the
false statement 7  3. Thus, the equation has
no solution.
The solution set is  .
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Objective 4: Example
4b.
Solve: 3  x  1  9  8 x  6  5 x
3  x  1  9  8 x  6  5 x
3x  3  9  3x  6
3x  6  3x  6
3x  3x  6  3x  3x  6
66
The original equation is equivalent 6  6, Which is
true for every value of x. The equation’s solution is
all real numbers or  x x is a real number.
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Objective 5
Solve applied problems using formulas.
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Linear Equations – An Application
5d
The formula p  15 
describes the pressure of
11
sea water p, in pounds per square foot, at a depth d
feet below the surface.
At what depth is the pressure 30 pounds per
square foot?
Substitute the given pressure into the equation for p,
p  30. The equation becomes:
5d
30  15 
11
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Linear Equations – An Application (cont)
5d
30  15 
11
5d 

30 11   15 
11

11 

Multiply both sides by 11
the LCD.
330  165  5d
Simplify.
330  165  165  165  5d Subtract 165 from both
sides.
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Linear Equations – An Application (cont)
165  5d
165 5d

5
5
d  33
Simplify.
Divide both sides by 5.
Simplify.
The depth where the pressure is 30 pounds per
square foot is 33 feet.
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Objective 5: Example
5.
It has been shown that the person’s with a
low sense of humor have higher levels of
depression in response to negative life
events than those with a high sense of
humor. This can be modeled by the following
formulas:
10
53
Low-Humor Group: D 
x
9
9
1
26
High-Humor Group: D  x 
9
9
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Objective 5: Example (cont)
Where x represents the intensity of a negative
life event (from a low of 1 to a high of 10) and D
is the low level of depression in response to that
event.
If the low-humor group averages a level of
depression of 10 in response to a negative life
event, what is the intensity of that event?
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Objective 5: Example (cont)
10
53
D
x
9
9
10
53
10 
x
9
9
53 
 10
9  10  9  x 

9
9


90  10 x  53
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Objective 5: Example (cont)
90  53  10 x  53  53
37  10 x
37 10 x

10 10
x  3.7
The formula indicates that if the low-humor group
averages a level of depression of 10 in response
to a negative life event, the intensity of that event
is 3.7
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