Download Lesson 2: Solving Equations and Inequalities

Lesson 2: Solving Equations and Inequalities
Problem Solving Activities solutions
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: 402 = 1600 , 412 = 1681 ,
422 = 1764 , 432 = 1849 , 442 = 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 452 = 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 x 2 , I will be x − 20 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 D1 and D2 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 D1 = 4t . It took the second man 3
hours more, so the time he walked is t + 3 hours. The distance he
walked is D2 = 3 ( t + 3) . Since the distances are the same, we
have:
Algebra 1
© 2009 Duke University Talent Identification Program
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Lesson 2: Solving Equations and Inequalities
Problem Solving Activities solutions
D1 = D2
4t = 3 ( t + 3)
4t = 3t + 9
t =9
The first man walked for 9 hours so his distance was
D1 = 4 ⋅ 9 = 36 kilometers .
The distance for the second man is
D2 = 3 ( 9 + 3) = 3 ⋅ 12 = 36 kilometers .
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
n = n − 1.8
5
2
Multiply both sides by the LCD (10) and we get:
2 
1

10  n  = 10  n − 1.8 
5 
2

4n = 5n − 18
−n = −18
n = 18
Check:
?
2
1
(18 ) = (18 ) − 1.8
5
2
?
7.2 = 9 − 1.8
?
7.2 = 7.2 Answer: The number is 18.
Algebra 1
© 2009 Duke University Talent Identification Program
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Lesson 2: Solving Equations and Inequalities
Problem Solving Activities solutions
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 x + 2 is the next consecutive
odd number, and x + 4 is the third consecutive odd number. The
equation we get is:
2x + 3 ( x + 4) = 5 ( x + 2) + 2
Solving this equation gives:
2x + 3 ( x + 4) = 5 ( x + 2) + 2
2 x + 3x + 12 = 5x + 10 + 2
5x + 12 = 5x + 12
0=0
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 s + 10 , and Marsha’s age will be
7s + 10 . Since Marsha’s age in 10 years is 3 times her son’s age
in 10 years, we get the equation:
7s + 10 = 3 ( s + 10 )
Solving this equation gives:
Algebra 1
© 2009 Duke University Talent Identification Program
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Lesson 2: Solving Equations and Inequalities
Problem Solving Activities solutions
7s + 10 = 3 ( s + 10 )
7s + 10 = 3s + 30
4s = 20
s = 5 years old
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.
Algebra 1
© 2009 Duke University Talent Identification Program
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