6'5 Translating Words into Algebraic Symbols

1
6.5 Translating Words into Algebraic Symbols
2
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.
3
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
4
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
5
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
6
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
7
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 .
8
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 .
9
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.
10
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.
11
6.6 Applications Involving Equations
12
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.
13
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
14
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
15
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.
16
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
17
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
18
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.

19
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.

20
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
21
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

22
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 ?
23
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

24
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.
25
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.
26
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
27
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
28
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.