words into symbols & story problems

8/7/2019 words into symbols & story problems

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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


Verbal Expression

4less than a number



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


Verbal Expression

14times a number



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


Verbal Expression

The quotient of 6 and a number



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


Step 4 Solve the equation obtained in Step 3.



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


Step 5 State the answer. The width of the rectangle is 15 ft and the length is

2 + 3(15) = 47 ft.



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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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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each piece be? Solve the equation.

1

9 (8) 1008

9 1 100

9 99

11

x

x

x

x

!

!

!

!


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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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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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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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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

!

!

!

!

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