Answered

Dec 23, 8:45:08 PM
Find three consecutive POSITIVE ODD integers such that when you multiply the first
and third the result is nine less than six times the second integer.

Answer :

Answer:

The three consecutive POSITIVE ODD integers are: 3, 5 and 7

Step-by-step explanation:

If x is an integer, then there consecutive positive odd integers are:

x , x + 2, x + 4 are three consecutive integers.

Where

First integer, a = x

Second integer, b = x + 2

Third integer, c = x + 4

We are told to:

Find three consecutive POSITIVE ODD integers such that when you multiply the first and third the result is nine less than six times the second integer.

Hence, this is represented as:

a × c = 6(b) - 9

Where

a = x

b = x + 2

c = x + 4

x (x + 4 ) = 6(x + 2) - 9

x ² + 4x = 6x + 12 - 9

x² +4x - 6x - 12 + 9= 0

x² - 2x - 3 = 0

Factorising

x² +x - 3x - 3 = 0

x (x + 1) -3(x + 1) = 0

(x - 3) (x + 1) = 0

x - 3 = 0

x = 3

x + 1 = 0

x = -1

Since, x cannot be negative for this question, x = 3

First integer, a = x = 3

Solving for:

Second integer, b = x + 2

b = 3 + 2 = 5

Solving for :

Third integer, c = x + 4

c = 3 + 4 = 7

Therefore, the three consecutive POSITIVE ODD integers are: 3, 5 and 7

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