Ch 3 Sec 3: Slide #1 Columbus State Community College Chapter 3 Section 3 Solving Application Problems with One Unknown Quantity

Ch 3 Sec 3: Slide #1

Columbus State Community College

Chapter 3 Section 3

Solving Application Problems with

One Unknown Quantity


Solving Application Problems with One Unknown Quantity

1. Translate word phrases into algebraic expressions.

2. Translate sentences into equations.

3. Solve application problems with one unknown quantity.


Translating Word Phrases into Algebraic Expressions

Write each phrase as an algebraic expression. Use x as the variable.

EXAMPLE 1 Translating Word Phrases

Words Algebraic Expression

a) A number plus 7 x + 7 or 7 + x

b) The sum of 3 and a number 3 + x or x + 3

c) 6 more than a number x + 6 or 6 + x

d) –15 added to a number –15 + x or x + –15

e) A number increased by 2 x + 2 or 2 + x

Two correct ways to write

each addition expression.






f) 8 less than a number x – 8

g) A number subtracted from 1 1 – x

h) 6 subtracted from a number x – 6

i) A number decreased by 4 x – 4

j) 9 minus a number 9 – x

Only one correct way to write

each subtraction expression.


The Order of Terms

Recall that addition can be done in any order, so x + 4 gives the same

result as 4 + x. This is not true in subtraction, so be careful. 3 – x does

not give the same result as x – 3.

CAUTION






a) 8 times a number 8x

b) The product of 32 and a number 32x

c) Double a number (meaning “2 times”) 2x

d) The quotient of –7 and a number

e) A number divided by 4

f) 14 subtracted from 3 times a number

–7x

x4

3x – 14






g) The result is =


Translating a Sentence into an Equation

If 4 times a number is added to 15, the result is 11. Find the number.

EXAMPLE 3 Translating a Sentence into an Equation

Let x represent the unknown number.

4x

4 times a number

+

added to

15

15

=

is

11

11

4x + 15 = 11

– 15 – 15

4x + 15 = –4

4 4

x + 15 = –1


Translating a Sentence into an Equation

If 4 times a number is added to 15, the result is 11. Find the number.

EXAMPLE 3 Translating Sentence into an Equation

4 • –1

If 4 times a number is added to 15, the result is 11.

+ 15 = 11

Check Go back to the words of the original problem.

Does 4 • –1 + 15 really equal 11?

Yes. 4 • –1 + 15 = –4 + 15 = 11.

So –1 is the correct solution because it “works” when you put it back

into the original problem.


Solving an Application Problem

Step 1 Read the problem once to see what it is about. Read it

carefully a second time. As you read, make a sketch or

write word phrases that identify the known and the

unknown parts of the problem.

Step 2 (a) If there is one unknown quantity, assign a variable to

represent it. Write down what your variable represents.

Step 2 (b) If there is more than one unknown quantity, assign a

variable to represent “the thing you know the least about.”

Then write variable expression(s), using the same variable,

to show the relationship of the other unknown quantities

to the first one. continued…



Step 3 Write an equation, using your sketch or word phrases as a

guide.

Step 4 Solve the equation.

Step 5 State the answer to the question in the problem and label

your answer.

Step 6 Check whether your answer fits all the facts given in the

original statement of the problem. If it does, you are done.

If it doesn’t, start again at Step 1.

Solving an Application Problem (continued)



Solving an Application Problem: One Unknown Quantity

EXAMPLE 4 Application Problems: One Unknown Quantity

Mike gained 8 pounds over the winter. He went on a diet and lost 12

pounds. Then he regained 15 pounds and weighed 192 pounds. How

much did he weigh originally?

Step 1 Read the problem once. It is about Mike’s weight.

Read it a second time and write word phrases.

Step 2(a) There is only one unknown quantity, so assign a variable

to represent it. Let w represent Mike’s original weight.

Step 3 Write an equation, using the phrases you wrote as a guide.

Start Weight

w

Lost 12

– 12

Gained 15

+ 15

Ending Weight

= 192

Gained 8

+ 8



w + 8 – 12 + 15 = 192

w + 11 = 192

– 11 – 11

w = 181







Step 5 State the answer to the question, “How much did he weigh

originally?” Mike originally weighed 181 pounds.







Step 6 Check the solution by going back to the original problem

and inserting the solution.

Because 181 pounds “works” when you put it back into the original

problem, you know it is the correct solution.

Originally weighed 181 pounds.

Gained 8 lbs, so 181 + 8 = 189 pounds

Lost 12 lbs, so 189 – 12 = 177 pounds

Gained 15 lbs, so 177 + 15 = 192 pounds







Joe purchased 6 boxes of calendars. He gave his secretary 12 calendars,

his business partner 5 calendars, and his accountant 19 calendars. If Joe

has 12 calendars remaining, how many were in each box he purchased?

Step 1 Read the problem once. It is about 6 boxes of calendars.

Unknown: the number of calendars in each box purchased.

Known: 6 boxes; gave away 12, 5, and 19 calendars.

Step 2(a) There is only one unknown quantity. Assign a variable, b,

to represent the number of calendars in each box.




Step 3 Write an equation.




Number of

boxes

6

Number in

each box

• b

Gave

away

– 12

Amount

remaining

= 12

Gave

away

– 5



Gave

away

– 19



6b – 12 – 5 – 19 = 12

6b – 36 = 12

+ 36 + 36

6b = 48

6 6

b = 8







Step 5 State the answer. There were 8 calendars in each box.

Step 6 Check the solution using the original problem.

6 boxes each containing 8 calendars, so 6 • 8 = 48 calendars.

Gave away 12, 5, and 19 calendars, so

48 – 12 – 5 – 19 = 12 calendars.

8 calendars in each box is the correct solution because it “works.”







Ty won 4 less than three times as many golf matches as Mike. If Ty

won 11 matches, how many matches did Mike win?

Step 1 Read the problem. It is about the number of golf matches

won by two people.

Step 2(a) Assign a variable, m, to represent the number of matches

that Mike won.

Step 3 Write an equation.

The number of matches

Ty won

11

4 less than three times the

number of matches Mike won

3m – 4

is

=

Solving More Complex Application Problems: One Unknown Quantity

EXAMPLE 6 Complex Application Problems: One Unknown


11 = 3m – 4

+ 4 + 4

15 = 3m

3 3

5 = m





Step 4 Solve the equation


Step 6 Check the solution using the original problem.

Three times the number of Mikes’s wins 3 • 5 = 15

Less 4 15 – 4 = 11

Ty won 11 matches.

The correct solution is: Mike won 5 golf matches.





Step 5 State the answer. Mike won 5 golf matches.


Solving Application Problems with One Unknown Quantity

Chapter 3 Section 3 – Completed

Written by John T. Wallace

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Ch 3 Sec 3: Slide #1 Columbus State Community College Chapter 3 Section 3 Solving Application Problems with One Unknown Quantity