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# Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 22?

### 13 Answers

- PinkgreenLv 73 weeks ago
Let the 3 odd numbers be

x-2, x, x+2

x-2+2x+3(x+2)=22

=>

6x+4=22

=>

6x=18

=>

x=3

=>

the numbers are 1,3,5.

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- Engr. RonaldLv 73 weeks ago
2n + 1 +2(2n + 3) + 3(2n + 5) = 22

2n + 1 + 4n + 6 + 6n + 15 = 22

12n + 22 = 22

12n = 22 - 22

12n = 0

n = 0

1st odd integer = 2n + 1 = 2(0) + 1 = 1 answer//

2nd odd integer = 2n + 3 = 2(0) + 3 = 3 answer//

3rd odd integer = 2n + 5 = 2(0) + 5 = 5 answer//

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- L. E. GantLv 73 weeks ago
the integers are "consecutive and odd" so call them

2n + 1, 2n+3 and 2n+5

hence

2n+1 + 4n+6 + 6n + 15 = 22

==> 12n + 22 = 22

==> n = 0

so

1, 3 and 5

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- MICHAEL KLv 74 weeks ago
the three consecutive odd integers are 1 and 3 and 5.

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- Φ² = Φ+1Lv 74 weeks ago
The average of these numbers is 22/6 = 3⅔ and this is going to fall between the second and third odd integers, so the three odd integers are 1, 3, and 5.

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- KrishnamurthyLv 74 weeks ago
Three consecutive odd integers such that

the sum of the first, two times the second,

and three times the third is 22:

(x - 2) + 2x + 3(x + 2) = 22

x = 3

Answer: 1, 3 and 5.

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- llafferLv 74 weeks ago
We want three consecutive odd integers. If we call the middle number "x", then the ones before and after it are (x - 2) and (x + 2).

The sum of the first, two times the second, and three times the third is 22:

(x - 2) + 2x + 3(x + 2) = 22

Simplify and solve for x:

x - 2 + 2x + 3x + 6 = 22

6x + 4 = 22

6x = 18

x = 3

That's your middle number, so the other two are: 1 and 5

Your three numbers are:

1, 3, and 5.

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- MorningfoxLv 74 weeks ago
The integers are 1, 3, and 5. Because 1 + 6 + 15 = 22.

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- 4 weeks ago
Your integers will be x - 2 , x , x + 2

(x - 2) + 2 * x + 3 * (x + 2) = 22

x - 2 + 2x + 3x + 6 = 22

6x + 4 = 22

6x = 18

x = 3

1 , 3 , 5

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