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Chapter 11
Algebra: Solving
Equations and
Problems.
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Algebra: Solving Equations and
Problems
CHAPTER
11
11.1 Introduction to Algebra
11.2 Solving Equations: The Addition Principle
11.3 Solving Equations: The Multiplication Principle
11.4 Using the Principles Together
11.5 Applications and Problem Solving
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11.5
Applications and Problem Solving
OBJECTIVES
a
b
Translate phrases to algebraic expressions.
Solve applied problems by translating to equations.
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In algebra, we translate problems to equations. The
different parts of an equation are translations of word
phrases to algebraic expressions.
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Key Words, Phrases, and Concepts
Addition (+)
Subtraction (–)
add
subtract
added to
subtracted from
sum
difference
total
minus
plus
less than
more than
decreased by
increased by
take away
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Key Words, Phrases, and Concepts
Multiplication ()
Division ()
multiply
divide
multiplied by
divided by
product
quotient
times
of
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Example
Translate to an algebraic expression:
1. The product of 4 and some number.
2. Twelve more than some number.
Solution
1. The product of 4 and some number
4x
4x
4x
2. Twelve more than some number.
x + 12
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Example
Translate to an algebraic expression:
Five more than a number.
Solution
Five more than a number.
x = the number
Five more than a number
5
+
x
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Example
Translate to an algebraic expression.
a. Eight more than some number
x + 8 or 8 + x
b. Four more than twice some number
4 + 2y or 2y + 4
c. The difference of two numbers
w–z
d. Nine less than the product of two numbers
xy – 9
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Five Steps for Problem Solving
1. Familiarize yourself with the problem
situation.
2. Translate to an equation.
3. Solve the equation.
4. Check your possible answer in the original
problem.
5. State the answer clearly.
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To familiarize yourself with a problem.
a. If a problem is given in words, read it carefully.
Reread the problem, perhaps aloud. Try to verbalize
the problem if you were explaining it to someone
else.
b. Choose a variable (or variables) to represent the
unknown and clearly state what the variable
represents. Be descriptive! For example, let L = the
length, d = the distance, and so on.
c. Make a drawing and label it with known
information, using specific units if given. Also,
indicate unknown information.
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d. Find further information. Look up formulas or
definitions with which you are not familiar.
(Geometric formulas appear on the inside back
cover of this text.) Consult a reference librarian or
an expert in the field.
e. Create a table that lists all the information you have
available. Look for patterns that may help in the
translation to an equation.
f. Think of a possible answer and check the guess.
Observe the manner in which the guess is checked.
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Example
A 480-in. piece of pipe is cut into two pieces. One
piece is three times the length of the other. Find the
length of each piece of pipe.
Solution
1. Familiarize. Make a drawing. Noting the lengths.
480 in
3x
x
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continued
2. Translate. From the statement of the problem.
One piece is three times the length of the other the
total is 480 inches.
x + 3x = 480
3. Solve.
x + 3x = 480
4x = 480
4
4
x = 120 inches
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continued
4. Check. Do we have an answer to the problem?
No, we need the lengths of both pieces of pipe.
If x = 120 the length of one piece
3x = the length of the other piece. 3(120) = 360
inches
Since 120 + 360 = 480 our answer checks.
5. State. One section of pipe is 120 inches and the
other section is 360 inches.
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Example
Digicon prints digital photos for $0.12 each plus $3.29 shipping
and handling. Your weekly budget for the school yearbook is
$22.00. How many prints can you have made if you have
$22.00?
Solution
1. Familarize. Suppose the yearbook staff takes 220
digital photos. Then the cost to print them would be
the shipping charge plus $0.12 times 220.
$3.29 + $0.12(220) which is $29.69. Our guess
of 220 is too large, but we have familiarized ourselves with
the way in which the calculation is made.
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continued
2. Translate.
Rewording: Shipping
Translating:
3. Carry out.

$3.29
plus
photo cost


is


0.12( x ) 
$22

22
3.29  0.12x  22
0.12x  18.71
x  155.9  155
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continued
4. Check. Check in the original problem.
$3.29 + 155(0.12) = $21.89, which is less than $22.00.
5. State. The yearbook staff can have 155 photos
printed per week.
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Example
You are constructing a triangular kite. The second
angle of the kite is three times as large as the first. The
third angle is 10 degrees more than the first. Find the
measure of each angle.
x + 10
Solution
3x
1. Familiarize. Make a drawing and
write in the given information.
x
2. Translate. To translate, we need to recall that the
sum of the measures of the angles in a triangle is
180 degrees.
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continued
2. Translate (continued).
Measure of
first angle

x
measure of
+


measure of
second angle
+
third angle

3x



x  10 

3. Carry out.
x  3x   x  10  180
5x  10  180
5x  170
x  34
The measures for the angles appear to be:
first angle:
x = 34
second angle: 3x = 3(34) = 102;
third angle: x + 10 = 34 + 10 = 44
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is


180

180
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continued
4. Check. Consider 34, 102 and 44 degrees. The sum of
these numbers is 180 degrees and the second angle is three
times the first angle. The third angle is 10 degrees more
than the first. These numbers check.
5. State. The measures of the angles are 34, 44 and 102
degrees.
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