Transcript Math 120 Chapter 1-1.4

Solving Linear Equations
1.4
Copyright © 2005 Pearson Education, Inc.
Definitions



Algebra: a generalized form of arithmetic.
Variables: used to represent numbers
Algebraic expression: a collection of variables,
numbers, parentheses, and operation symbols.

Examples:
4x  2
x, x  4, 4(3 y  5),
, y 2  8y  2
3x  5
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Slide 6-2
Order of Operations

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1. First, perform all operations within
parentheses or other grouping symbols
(according to the following order).
2. Next, perform all exponential operations
(that is, raising to powers or finding roots).
3. Next, perform all multiplication and
divisions from left to right.
4. Finally, perform all additions and
subtractions from left to right.
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Slide 6-3
Example: Evaluating an Expression

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Evaluate the expression x2 + 4x + 5 for x = 3.
Solution:
x2 + 4x + 5
= 32 + 4(3) + 5
= 9 + 12 + 5
= 26
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Slide 6-4
Example: Substituting for Two Variables
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Evaluate 4 x 2  3 xy  5y 2 when x = 3 and y = 4.
Solution:
2
2
4 x  3 xy  5 y
 4(3)2  3(3)(4)  5(4 2 )
 4(9)  36  5(16)
 36  36  80
 0  80
 80
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Slide 6-5
Linear Equations
in
One Variable
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Definitions

Like terms are terms that have the same
variables with the same exponents on the
variables.
2
2
2x, 7x
 5x , 8x

Unlike terms have different variables or different
exponents on the variables.
2x, 7
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 5x , 6x
3
2
Slide 6-7
Properties of the Real Numbers
Equality Principles
a(b + c) = ab + ac
Distributive property
a+b=b+a
Commutative property of
addition
ab = ba
Commutative property of
multiplication
(a + b) + c = a + (b + c)
Associative property of
addition
(ab)c = a(bc)
Associative property of
multiplication
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Slide 6-8
Example: Combine Like Terms
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8x + 4x
= (8 + 4)x
= 12x
5y  6y
= (5  6)y
= y
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

x + 15  5x + 9
= (1 5)x + (15+9)
= 4x + 24
3x + 2 + 6y  4 + 7x
= (3 + 7)x + 6y + (2  4)
= 10x + 6y  2
Slide 6-9
Solving Equations
Addition Property of Equality
If a = b, then a + c = b + c for
all real numbers a, b, and c.
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
Find the solution to the
equation
x  9 = 24.
x  9 + 9 = 24 + 9
x = 33
Check: x  9 = 24
33  9 = 24 ?
24 = 24 true
Slide 6-10
Solving Equations continued
Subtraction Property of
Equality
If a = b, then a  c = b  c
for all real numbers a, b,
and c.
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
Find the solution to the
equation
x + 12 = 31.
x + 12  12 = 31  12
x = 19
Check: x + 12 = 31
19 + 12 = 31 ?
31 = 31 true
Slide 6-11
Solving Equations continued
Multiplication Property of
Equality
If a = b, then a • c = b • c for
all real numbers a, b, and c,
where c  0.

Find the solution to the
equation x
 9.
7
x
9
7
x
7     7(9)
7
1
7x
 63
1
7
x  63
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Slide 6-12
Solving Equations continued
Division Property of Equality
a b
If a = b, then c  c for all
real numbers a, b, and c,
c  0.
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
Find the solution to the
equation 4x = 48.
4 x  48
4 x 48

4
4
x  12
Slide 6-13
General Procedure for Solving Linear
Equations
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If the equation contains fractions, multiply both sides of
the equation by the lowest common denominator (or
least common multiple). This step will eliminate all
fractions from the equation.
Use the distributive property to remove parentheses
when necessary.
Combine like terms on the same side of the equal sign
when possible.
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Slide 6-14
General Procedure for Solving Linear
Equations continued

Use the addition or subtraction property to
collect all terms with a variable on one side of
the equal sign and all constants on the other
side of the equal sign. It may be necessary to
use the addition or subtraction property more
than once. This process will eventually result in
an equation of the form ax = b, where a and b
are real numbers.
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Slide 6-15
General Procedure for Solving Linear
Equations continued

Solve for the variable using the division or
multiplication property. This will result in an
answer in the form x = c, where c is a real
number.
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Slide 6-16
Example: Solving Equations

Solve 3x  4 = 17.
3 x  4  17
3 x  4  4  17  4
3 x  21
3 x 21

3
3
x 7
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Slide 6-17
Solve 21 = 6 + 3(x + 2)

21  6  3( x  2)
21  6  3 x  6
21  3 x  12
21  12  3 x  12  12
9  3x
9 3x

3 3
3x
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Slide 6-18
Solve 8x + 3 = 6x + 21

8 x  3  6 x  21
8 x  3  3  6 x  21  3
8x
8x  6x
2x
2x
2
x
 6 x  18
 6 x  6 x  18
 18
18

2
9
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Slide 6-19
Solve 6(x  2) + 2x + 3 = 4(2x  3) + 2

6( x  2)  2 x  3  4(2 x  3)  2
6 x  12  2 x  3  8 x  12  2
8 x  9  8 x  10
8 x  8 x  9  8 x  8 x  10
9  10

False, the equation has no solution. The
equation is inconsistent.
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Slide 6-20
Solve 4(x + 1)  6(x + 2) = 2(x + 4)

4( x  1)  6( x  2)  2( x  4)
4 x  4  6 x  12  2 x  8
2 x  8  2 x  8
2 x  2 x  8  2 x  2 x  8
8  8
8  8  8  8
00

True, 0 = 0 the solution is all real numbers.
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Slide 6-21
Proportions

A proportion is a statement of equality between
two ratios.

Cross Multiplication

a c
If  , then ad = bc, b  0, d  0.
b d
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Slide 6-22
To Solve Application Problems Using
Proportions
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Represent the unknown quantity by a variable.
Set up the proportion by listing the given ratio on the
left-hand side of the equal sign and the unknown and
other given quantity on the right-hand side of the equal
sign. When setting up the right-hand side of the
proportion, the same respective quantities should
occupy the same respective positions on the left and
right.
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Slide 6-23
To Solve Application Problems Using
Proportions continued

For example, an acceptable proportion might be
miles miles

hour
hour
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Once the proportion is properly written, drop the
units and use cross multiplication to solve the
equation.
Answer the question or questions asked.
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Slide 6-24
Example

A 50 pound bag of
fertilizer will cover an
area of 15,000 ft2. How
many pounds are needed
to cover an area of
226,000 ft2?

11,300,000  15,000 x
11,300,000 15,000 x

15,000
15,000
753.33  x

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50 pounds
x

2
15,000 ft
226,000 ft 2
(50)(226,000)  15,000 x
754 pounds of fertilizer would
be needed.
Slide 6-25
Translating Words to Expressions
Phrase
Mathematical Expression
Ten more than a number
x + 10
Five less than a number
x5
Twice a number
2x
Eight more than twice a number
2x + 8
The sum of three times a
number decreased by 9.
3x  9
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Slide 6-26
To Solve a Word Problem



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Read the problem carefully at least twice to be sure that
you understand it.
If possible, draw a sketch to help visualize the problem.
Determine which quantity you are being asked to find.
Choose a letter to represent this unknown quantity.
Write down exactly what this letter represents.
Write the word problem as an equation.
Solve the equation for the unknown quantity.
Answer the question or questions asked.
Check the solution.
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Slide 6-27
Example
The bill (parts and labor) for the repairs of a car
was $496.50. The cost of the parts was $339.
The cost of the labor was $45 per hour. How
many hours were billed?


Let h = the number of hours billed
Cost of parts + labor = total amount
339 + 45h = 496.50
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Slide 6-28
Example continued

339  45h  496.50
339  339  45h  496.50  339
45h  157.50
45h 157.50

45
45
h  3.5

The car was worked on for 3.5 hours.
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Slide 6-29
Example

Sandra Cone wants to fence in a rectangular
region in her backyard for her lambs. She only
has 184 feet of fencing to use for the perimeter
of the region. What should the dimensions of
the region be if she wants the length to be 8 feet
greater than the width?
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Slide 6-30
continued, 184 feet of fencing, length 8
feet longer than width
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Let x = width of region
Let x + 8 = length
P = 2l + 2w
x
x+8
184  2( x )  2( x  8)
184  2 x  2 x  16
184  4 x  16
168  4 x
42  x
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The width of the region is 42 feet
and the length is 50 feet.
Slide 6-31