Find three consecutive odd integers such that the sum of the smallest and 7 times the largest is 68
2 answers:
If you are trying to find the equation to this problem, I am pretty sure that it is 2x + x + 4 = 25.
Please mark as brainliest
First we must assign variables to the integers.
x = smallest
y = middle
z = largest
x + 7z = 68 ← This equation represents the given information
We can assume that "z" is less than 11 because 7*11 = 77, which is greater than 68.
Our possible consecutive odd integers could be:
a) 1, 3, 5
b) 3, 5, 7
c) 5,7, 9
Lets to plug in the "x" and "z" values
a) 1 + 7(5) = 1+ 35 = 36 This is incorrect, because 36 ≠ 68
b) 3 + 7(7) = 3 + 49 = 52 This is incorrect as well, because 52 ≠ 68
That leaves our final choice. Let's check to be certain
c) 5 + 7(9) = 5 + 63 = 68
5, 7, 9 is the 3 consecutive integers such that the sum of the smallest and 7 times the largest is 68.
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It pretty much would be 400