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6.5 Translating Words into Algebraic Symbols

In order to solve application problems, it is

necessary to translate English phrases into

algebraic symbols. The following are some common

phrases and their mathematic translation.

Applications

- Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are

numbers)

Verbal Expression

Addition

The sum of a number and 2

x 2

3 more than a number

x 3

7 plus a number

7 x

16 added to a number

x 16

A number increased by 9

x 9

The sum of two numbers

x y

Applications

- Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are

numbers)

Verbal Expression

Subtraction

4 less than a number

x 4

10 minus a number

10 x

A number decreased by 5

x 5

A number subtracted from 12

12 x

The difference between two numbers

x y

Applications

- Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are

numbers)

Verbal Expression

Multiplication

14 times a number

14x

A number multiplied by 8

8x

3x

Triple (three times) a number

xy

The product of two numbers

Applications

- Translating from Words to Mathematical Expressions

Mathematical Expression (where x and y are

numbers)

Verbal Expression

Division

The quotient of 6 and a number

A number divided by 15

half a number

Applications

Caution

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, not 5 x. A number subtracted from 12

is expressed as 12 x, not x 12. For

division, the number by which we are dividing is

the denominator, and the number into 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

of x and y is translated as .

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 .

Applications

Translating Words into Equations

Verbal Sentence

Equation

Twice a number, decreased by 4, is 32.

2x 4 32

16x 25 87

If the product of a number and 16 is decreased by

25, the result is 87.

48

The quotient of a number and the number plus 6

is 48.

x 54

The quotient of a number and 8, plus the number,

is 54.

Applications

Distinguishing between Expressions and Equations

Decide whether each is an expression or an

equation.

(a)

4(6 x) 2x 1

There is no equals sign, so this is an expression.

(b)

4(6 x) 2x 1 15

Because of the equals sign, this is an equation.

Note that the expression in part (a) simplifies

to the expression 2x 23 and the equation in

part (b) has solution 19.

6.6 Applications Involving Equations

Applications

Six Steps to Solving Application Problems

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

Solving a Geometry Problem

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 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.

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.

Let L 2 3W and P 124.

124 2(2 3W) 2W

Applications

Solving a Geometry Problem

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 4

Solve the equation obtained in Step 3.

124 2(2 3W) 2W

124 4 6W 2W

Remove parentheses

124 4 8W

Combine like terms.

124 4 4 8W 4

Subtract 4.

120 8W

120 8W

Divide by 8.

8 8

15 W

Applications

Solving a Geometry Problem

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 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.

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 the measures

are mixed ft and in convert to in. - Step 2 draw a picture.

100 in

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

x

x

x

x

x

x

x

x

x

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.

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.

Distribute 1000 into 3 parts so that one part

will three 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. Since there are 3 unknowns we

need 2 more expressions using the variable . - Write an equation

x

3x

3x x

Distribute 1000 into 3 parts so that one part

will three times as large as the second and the

third part will be as large as the sum of the

other two.

- Solve the equation

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 1

Read the problem. We must find the amount

invested in each account.

Step 2

Assign a variable.

The formula for interest is I p r t.

Let x the amount to invest at 3

50,000 x the amount to invest at 5.

Rate (as a decimal)

Interest

Principle

Time

.03

0.03x

x

1

.05

50,000 x

1

.05(50,000 x)

50,000

2180

Totals

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 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

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 2500 .05x 2180

Multiply.

.02x 2500 2180

Combine like terms.

.02x 320

Subtract 2500

x 16,000

Divide by .02.

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.

Step 6

Check by finding the annual interest at each

rate they should total 2180.

and

0.03(16,000) 480

.05(34,000) 1700

480 1700 2180, as required.

Solving a Mixture Problem

A chemist must mix 12 L of a 30 acid solution

with some 80 solution to get a 60 solution. How

much of the 80 solution should be used

- EXAMPLE 7

Step 1

Read the problem. The problem asks for the amount

of 80 solution to be used.

Step 2

Assign a variable. Let x the number of liters

of 80 solution to be used.

80

30

80

30

12 L

Unknown number of liters, x

(12 x)L

A chemist must mix 12 L of a 30 acid solution

with 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)L

A chemist must mix 12 L of a 30 acid solution

with some 80 solution to get a 60 solution. How

much of the 80 solution should be used

- Solve the equation.

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