words into symbols & story problems
TRANSCRIPT
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6.5 Translating Words into
Algebraic Symbols
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In order to solve application
problems, it is necessary totranslate English phrases into
algebraic symbols. The followingare some common phrases and
their mathematic translation.
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Applications
Translating from Words to Mathematical Expressions
Verbal Expression
The sum of a number and 2
Mathematical Expression
(where xand y are numbers)
Addition
3more than a number
7 plus a number
16 added to a number
A numberincreased by 9
The sum of two numbers
x + 2
x + 3
7 + x
x + 16
x + 9
x + y
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Applications
Translating from Words to Mathematical Expressions
Verbal Expression
4less than a number
Mathematical Expression
(where xand y are numbers)
Subtraction
10 minus a number
A numberdecreased by 5
A numbersubtracted from 12
The difference between two
numbers
x 4
10 x
x 5
12 x
x y
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Applications
Translating from Words to Mathematical Expressions
Verbal Expression
14times a number
Mathematical Expression
(where xand y are numbers)
Multiplication
A numbermultiplied by 8
Triple (three times) a number
The product of two numbers
14x
8x
3x
xy
ofa number (used with
fractions and percent)
34
x34
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Applications
Translating from Words to Mathematical Expressions
Verbal Expression
The quotient of 6 and a number
Mathematical Expression
(where xand y are numbers)
Division
A numberdivided by 15
half a number
(x 0)6x
x
15
2
x
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CAUTION
Because subtraction and division are not commutative operations, be careful
to correctly translate expressions involving them. For example, 5 less than a
number is translated as x 5, not5 x. A number subtracted from 12 is
expressed as 12 x, notx 12.
For division, the numberbywhich we are dividing is the denominator, and
the numberinto which we are dividing is the numerator. For example, a
number divided by 15 and 15 divided into x both translate as . Similarly,
the quotient ofx and y is translated as .
Applications
Caution
x
15x
y
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Applications
Indicator Words for Equality
Equality
The symbol for equality, =, is often indicated by the word is. In fact, any
words that indicate the idea of sameness translate to =.
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Applications
Translating Words into Equations
Verbal Sentence Equation
16x 25=
87
If the product of a number and 16 is decreasedby 25, the result is 87.
= 48The quotient of a number and the number plus
6 is 48.x + 6
x
+ x = 54The quotient of a number and 8, plus the
number, is 54.8
x
Twice a number, decreased by 4, is 32. 2x 4= 32
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Applications
Distinguishing between Expressions
and Equations
(a) 4(6 x) + 2x 1
(b) 4(6 x) + 2x 1 = 15
There is no equals sign, so this is an expression.
Because of the equals sign, this is an equation.
Decide whether each is an expression or an equation.
Note that the expression in part (a) simplifies to the expression 2x + 23
and the equation in part (b) has solution 19.
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6.6 Applications Involving
Equations
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Six Steps to Solving Application Problems
Step 1 Read the problem, several times if necessary, until you understand
what is given and what is to be found.
Step 2 If possible draw a picture or diagram to help visualize the problem.
Step 3 Assign a variable to represent the unknown value, using diagrams
or tables as needed. Write down what the variable represents.
Express any other unknown values in terms of the variable.
Step 4 Write an equation using the variable expression(s).
Step 5 Solve the equation.
Step 6 Check the answer in the words of the original problem.
Applications
Six Steps to Solving Application Problems
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Applications
Solving a Geometry Problem
Step 1 Read the problem. We must find the length and width of the rectangle.
The length is 2 ft more than three times the width and the perimeter is
124 ft.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 2 Assign a variable. Let W= the width; then 2 + 3W= length.
Make a sketch.
W
2 + 3W
Step 3 Write an equation. The perimeter of a rectangle is given by the
formula P= 2L + 2W.
124 = 2(2 + 3W) + 2W Let L= 2 + 3Wand P= 124.
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Applications
Solving a Geometry Problem
Step 4 Solve the equation obtained in Step 3.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
124 = 2(2 + 3W) + 2W
124 = 4 + 6W+ 2W
124 4 = 4 + 8W 4
120 8W8 8
15 = W
Remove parentheses
124 = 4 + 8W Combine like terms.
Subtract 4.
Divide by 8.
120 = 8W
=
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Applications
Solving a Geometry Problem
Step 5 State the answer. The width of the rectangle is 15 ft and the length is
2 + 3(15) = 47 ft.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 6 Check the answer by substituting these dimensions into the words of
the original problem.
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Saw a board 8 ft 4 in into nine equal pieces.If the loss per cut is 1/8 in, how long will
each piece be? Step 1: the board is to cut into 9 equal parts
with 1/8 in wasted each cut. Since themeasures are mixed ft and in convert to in.
Step 2: draw a picture.
100 in
1
8
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Saw a board 8 ft 4 in into nine equal pieces.If the loss per cut is 1/8 in, how long will
each piece be? Assign a variable for the
unknown.
Let x = the length of each
equal piece.
Write an equation:
100 in
1
8
x x x x x x x x x
19 (8) 100
8
x !
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Saw a board 8 ft 4 in into nine equal pieces.If the loss per cut is 1/8 in, how long will
each piece be? Solve the equation.
1
9 (8) 1008
9 1 100
9 99
11
x
x
x
x
!
!
!
!
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Saw a board 8 ft 4 in into nine equal pieces.If the loss per cut is 1/8 in, how long will
each piece be? Each piece should be 11 in long.
Check in the problem.
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Distribute $1000 into 3 parts so that one part willthree times as large as the second and the third
part will be as large as the sum of the other two. Read carefully.
Make a table:
Assign a variable. Sincethere are 3 unknowns weneed 2 more expressions
using the variable .
Write an equation
Part 1 Part 2 Part 3
3(?) ?
3(?) + ?x3x 3x + x
3 3 1000x x x x !
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Distribute $1000 into 3 parts so that one part willthree times as large as the second and the third
part will be as large as the sum of the other two. Solve the equation
3 3 1000
8 1000
125
3 375
3 500
x x x x
x
x
x
x x
!
!
!
!
!
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Applications of Linear Equations
Solving an Investment Problem
Step 1 Read the problem. We must find the amount invested in each account.
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 2 Assign a variable.
Let x = the amount to invest at 3%;
50,000 x = the amount to invest at 5%.
Rate (as a decimal) Interest
.03 0.03x
Principle
.05
x
50,000 x
The formula for interest is I = p r t.
Time
1
1 .05(50,000 x)
50,000 2180 Totals
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Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Rate (as a decimal) Interest
.03 0.03x
Principle
.05
x
50,000 x
Time
1
1 .05(50,000 x)
50,000 2180 Totals
Step 3 Write an equation. The last column of the table gives the equation.
interest at 3% interest at 5% = total interest+
.03x .05(50,000 x) = 2180+
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Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 4 Solve the equation. We do so without clearing decimals.
.03x + .05(50,000) .05x = 2180 Distributive property
.03x .05(50,000 x) = 2180+
.03x + 2500
.0
5x = 21
80 Multiply.
.02x + 2500 = 2180 Combine like terms.
.02x = 320 Subtract 2500
x = 16,000 Divide by .02.
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Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 5 State the answer. The company will invest $16,000 at 3%. At 5%, the
company will invest $50,000 $16,000 = $34,000.
and
Step 6 Check by finding the annual interest at each rate; they should total
$2180.
0.03($16,000) = $480 .05($34,000) = $1700
$480 + $1700= $2180, as required.
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EXAMPLE 7
Solving a Mixture Problem
Step 1 Read the problem. The problem asks for the amount of 80% solution
to be used.
A chemist must mix 12 L of a 30% acid solution with some 80% solution to geta 60% solution. How much of the 80% solution should be used?
Step 2 Assign a variable. Let x = the number of liters of 80% solution to be
used.
+ =30% 80% 30%
80%
12 L Unknown
number of liters, x
(12 + x)L
60%
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A chemist must mix 12 L of a 30% acid solutionwith some 80% solution to get a 60% solution.
How much of the 80% solution should be used?
Write an equation.
+ =
30% 80%12 L x (12 + x)L60%
.30(12) .80 .60(12 )x x !
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A chemist must mix 12 L of a 30% acid solutionwith some 80% solution to get a 60% solution.
How much of the 80% solution should be used?
Solve the equation.
.30(12) .80 .60(12 )
3.6 .8 7.2 .6
.2 3.6
18
x x
x x
x
x
!
!
!
!