Beginning Algebra, 11th Edition

Solving a Consecutive Integer Problem

The product of the second and third of three consecutive integers is 2 more than 7

times the first integer. Find the integers.

Step 1 Readthe problem. Note that the integers are consecutive.

Step 2 Assign a variable.

Let

Then

and

Step 3 Write an equation.

The product of the second and third is 2 more than 7 times the first.

Step 4 Solve. Multiply.

Standard form Factor. Zero-factor property

Step 5 State the answer.The solutions 0 and 4 each lead to a correct answer.

0, 1, 2 or 4, 5, 6

Step 6 Check.The product of the second and third integers must equal 2 more than

7 times the first. Since and both sets

of consecutive integers satisfy the statement of the problem.

NOW TRY

1 # 2 = 7 # 0 + 2 5 # 6 = 7 # 4 + 2,

x= 0 or x= 4

x 1 x- 42 = 0

x^2 - 4 x= 0

x^2 + 3 x+ 2 = 7 x+ 2

1 x+ 121 x+ 22 = 7 x+ 2

x+ 2 = the third integer.

x+ 1 = the second integer,

x= the first integer.

EXAMPLE 2

402 CHAPTER 6 Factoring and Applications

OBJECTIVE 3 Solve problems by applying the Pythagorean theorem.

Applying the Pythagorean Theorem

Patricia Walker and Ali Ulku leave their office, Patricia traveling north and Ali trav-

eling east. When Ali is 1 mi farther than Patricia from the office, the distance be-

tween them is 2 mi more than Patricia’s distance from the office. Find their distances

from the office and the distance between them.

Step 1 Readthe problem again. There will be three answers to this problem.

EXAMPLE 3

Pythagorean Theorem

If a right triangle has longest side of length cand two other sides of lengths aand

b, then

The longest side, the hypotenuse,is opposite the

right angle. The two shorter sides are the legsof

the triangle.

a^2 b^2 c^2. Hypotenuse

Leg a c

Leg b

90

NOW TRY
EXERCISE 2
The product of the first and
second of three consecutive
integers is 2 more than 8
times the third integer. Find
the integers.

NOW TRY ANSWER

9, 10, 11 or -2, -1, 0

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