Lesson 2: Solving Equations and Inequalities Problem Solving Activities solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 1 of 4

I was x years old in the year x2.

~ Augustus de Morgan

1. The quotation above is de Morgan’s response when someone asked him

his age.

a. Augustus de Morgan lived during the 19th century. In what year was he born?

We start by squaring values for x: 240 1600= , 241 1681= , 242 1764= , 243 1849= , 244 1936= . Since 1849 is the

only one in the 19th century, de Morgan was 43 years old

in the year 1849. This means he was born in 1849 – 43 =

1806.

Answer: 1806

b. A person born in 1980 could say the same thing that de Morgan

said (since he/she will turn 45 in the year 245 2025= ). Find a way

to express your age that you will be in a certain year in terms of x.

Answers will vary. Sample answer – a student born in

2000 could say, “In the year 2x , I will be 20x − years

old.”

2. Two men start walking at the same time. One man walks at a rate of 4

kilometers per hour and the other walks at a rate of 3 kilometers per

hour. The first man arrives at the destination 3 hours before the second

man. What is the distance that the men walked?

Let 1D and 2D be the distance traveled by the first man and the

second man, respectively. If the first man walked t hours, then

the distance he walked is 1 4D t= . It took the second man 3

hours more, so the time he walked is 3t + hours. The distance he

walked is ( )2 3 3D t= + . Since the distances are the same, we

have:

Lesson 2: Solving Equations and Inequalities Problem Solving Activities solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 2 of 4

( )

1 2

4 3 3

4 3 9

9

D D

t t

t t

t

=

= +

= +

=

The first man walked for 9 hours so his distance was

1 4 9 36 kilometersD = ⋅ = .

The distance for the second man is

( )2 3 9 3 3 12 36 kilometersD = + = ⋅ = .

Answer: The men walked 36 kilometers.

3. Two-fifths of a number is 1.8 less than half the number. Find the

number.

2 1

1.85 2n n= −

Multiply both sides by the LCD (10) and we get:

2 110 10 1.8

5 2

4 5 18

18

18

n n

n n

n

n

= −

= −

− = −

=

Check:

( ) ( )?

?

?

2 118 18 1.8

5 2

7.2 9 1.8

7.2 7.2

= −

= −

= �

Answer: The number is 18.

Lesson 2: Solving Equations and Inequalities Problem Solving Activities solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 3 of 4

4. Find three consecutive odd numbers such that the sum of twice the first

number and three times the third number is 2 more than five times the middle number.

If x is the first odd number, then 2x + is the next consecutive odd number, and 4x + is the third consecutive odd number. The

equation we get is:

( ) ( )2 3 4 5 2 2x x x+ + = + +

Solving this equation gives:

( ) ( )2 3 4 5 2 2

2 3 12 5 10 2

5 12 5 12

0 0

x x x

x x x

x x

+ + = + +

+ + = + +

+ = +

=

This equation is true for every value of x.

Answer: For any three consecutive odd numbers, the given

equation holds. [NOTE: This equation also holds for 3

consecutive even integers.]

5. Marsha is 7 times older than her son. In 10 years, she’ll be 3 times as

old as her son. Find their ages.

Let s be the son’s age. Then Marsha’s age is 7s .

In 10 years, the son’s age will be 10s + , and Marsha’s age will be

7 10s + . Since Marsha’s age in 10 years is 3 times her son’s age

in 10 years, we get the equation:

( )7 10 3 10s s+ = +

Solving this equation gives:

Lesson 2: Solving Equations and Inequalities Problem Solving Activities solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 4 of 4

( )7 10 3 10

7 10 3 30

4 20

5 years old

s s

s s

s

s

+ = +

+ = +

=

=

The son is 5 years old. Marsha is 7 5 35⋅ = years old.

Check: In 10 years, son will be 15, Marsha will be 45.

Since 45 3 15= ⋅ , our answer checks.

Answer: The son is 5 years old; Marsha is 35 years old.