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Find three consecutive odd integers where the sum of the two largest is seven less than three times the smallest.

User Lani
by
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1 Answer

2 votes

Answer:

13, 15, 17

Explanation:

x - an integer then 2x + 1 is an odd number (the smallest one)

consecutive odd numbers increase by 2

so the next odd number (the middle number) is:

2x + 1 + 2 = 2x + 3

and the third (the largest) consecutive is:

2x + 3 + 2 = 2x + 5

the sum of the the two largest numbers is:

2x + 3 + 2x + 5

3 times the smallest number is:

3(2x + 1)

the sum of the two largest is seven less than three times the smallest, so:

2x + 3 + 2x + 5 = 3(2x + 1) - 7

4x + 8 = 6x + 3 - 7

-6x -6x

-2x + 8 = -4

-8 -8

-2x = -12

÷(-2) ÷(-2)

x = 6

2x+1 = 2•6+1 = 13

2x+3 = 2•6+3 = 15

2x+5 = 2•6+5 = 17

Check:

2x+3+2x+5 = 15+3+15+5 = 38

38+7 = 45

45÷3 = 15

User Manolete
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