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Answer :
Final answer:
To solve for three consecutive odd integers that satisfy the given condition, we express them as x, x + 2, and x + 4, resulting in the integers 9, 11, and 13.
Explanation:
To find three consecutive odd integers where eight more than the sum of the first two is equal to eleven less than three times the third, we must first let the smallest odd integer be 'x'. Because these are consecutive odd integers, the next two will be 'x + 2' and 'x + 4'.
The equation based on the given condition is:
x + (x + 2) + 8 = 3(x + 4) - 11
Combining like terms and solving for 'x' we get:
2x + 10 = 3x + 12 - 11
2x + 10 = 3x + 1
x = 9
So the three consecutive odd integers are 9, 11, and 13.
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The consecutive odd integers are 9, 11, 13
What are Consecutive odd integers?
Consecutive odd integers are odd integers that follow each other and they differ by 2. If x is an odd integer, then x + 2, x + 4 and x + 6 are consecutive odd integers.
Given that, three consecutive odd integers such that eight more than the sum of the first two is equal to eleven less then three times the third
Consecutive odd integers are separated by two.
Therefore, x, x+2 and x+4
8 more than the sum of the 1st two:
x + x + 2 + 8
equals 11 less than three times the third:
= 3(x+4) - 11
Therefore,
2x + 10 = 3(x+4) - 11
2x + 10 = 3x +12 - 11
2x + 10 = 3x + 1
10 = x + 1
9 = x
Hence, The numbers are 9, 11, 13
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