Download Section 2.5 An Introduction to Problem Solving

Section 2.5 An Introduction to Problem Solving
Algebraic Translations of English Phrases
English Phrases
Algebraic Expression
Addition
The sum of a number and 3 (The sum of 3 and a number)
Three more than a number
A number plus 3 (3 plus a number)
A number increased by 3
Three added to a number
Subtraction
A number minus 4
Four minus a number
A number decreased by 4
The difference between a number and 4
The difference between 4 and a number
Four fewer than a number
A number subtracted from 4
Four subtracted from a number
Multiplication
Five times a number
The product of 5 and a number (the product of a number and 5)
Two-thirds of a number
Five multiplied by a number (A number multiplied by 5)
Twice a number
3
3
3
3
3
4
4
4
4
4
4
4
4
5
5
2
3
5
2
Division
A number divided by 6 (the quotient of a number and 6)
6
6
The quotient of 6 and a number
1
The reciprocal of a number
More than one operation
The sum of twice a number and 5
Twice the sum of a number and 5
Five times the sum of 5 and twice a number
Five subtracted from 7 times a number
Seven times a number; increased by 5
Cheon-Sig Lee
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2
2
5 5
7
7
5
5
2
5
5
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Section 2.5 An Introduction to Problem Solving
Exercises
1.
(Solution 1)
A number increased by 10 is equal to 340
x + 10 = 340
Finding the number
10 340
10
10
330
2.
(Solution 2)
x is decreased by 7, the result is 21
x – 7 = 21
Solving the equation for x.
7 21
7
7
28
3.
(Solution 3)
The product of 8 and a number is 144.
Finding the number
8
144
8
144
8
8
18
8x = 144
4.
(Solution 4)
The quotient of the number and 15 is 3
15
Cheon-Sig Lee
3
Finding the number
15
15 ∙
3
3
15 ∙ 15
45
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Section 2.5 An Introduction to Problem Solving
5.
(Solution 5)
Three subtracted from the product of 2 and a number is 19.
2 3
Finding the number
2
3 19
3
3
2
22
2
22
2
2
11
19
(cf.) The product of 2 and a number subtracted from three is 19.
3 2
19
6.
(Solution 6)
Twice the sum of five and a number is 34.
2 5
34
Finding the number.
2 5
34
34
2 5
2
2
5
17
5
5
12
7.
(Solution 7)
Step 1: Translating English sentences to algebraic expressions.
$240
$240perweek ⇒
$0.25permile ⇒
$0.25
Step 2: Setting the equation: set x for the question.
The question is finding ‘the number of miles you can
travel.’ Thus let x be the number of miles.
$240
$0.25
∙ 1
∙
$375
Cheon-Sig Lee
Step 3: Solving the resulting equation for x.
240 0.25
375
240
240
0.25
135
0.25
135
0.25 0.25
540
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Section 2.5 An Introduction to Problem Solving
8.
(Solution 8)
Step 1: Translating English sentences to algebraic expressions.
$1510permonth ⟹ ⇒
.
$20
$20peryear ⟹ Step 2: Setting the equation: set x for the equation.
The question is finding years. Thus, let x be years after 2006.
$20
∙
$1810
$1510
Step 3: Solve the resulting equation for x.
1510 20
1810
1510 1510
20
300
20
300
20
20
15
Therefore, 15 years after 2006 the mortgage payments average will be $1810.
In 2021, the mortgage payments average will be $1810. ⟸ 2006 15 2021
9.
(Solution 9)
Step 1: Translating English sentences to algebraic expressions.
Let x be the height of the bookcase. Then the length of
the bookcase is
because the length is 3 times the height.
Step 2: Setting the equation.
The total lumber = 45.
3∙
4∙
45 3
4
45 3
12
Cheon-Sig Lee
Step 3: Solving the resulting equation.
3
12
45
15
45
15
45
15 15
3
Therefore, the height is 3 feet and the 9 feet.
3∙
3∙
3∙3 9
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Section 2.5 An Introduction to Problem Solving
10.
(Solution 10)
Method 1:
Step 1: Translating English sentences to algebraic expressions.
Reduction = Subtraction
The sale price $68 is 10% of the original price because
% 100% %
Step 2: Setting the equation.
Let x be the original price of the new sofa on sale. Then we have
$68
% %
$68
. ∙
.
←
%
Step 3: Solving the resulting equation for x.
68
. ∙
68 0.1
0.1 0.1
680
Method 2:
Let x be the original price of the new sofa. Then,
90%
90%
0.9 ∙ .
←
%
%
.
68
68
68
0.1
680
Cheon-Sig Lee
0.9 ∙
0.1
0.1
0.1
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