Answered

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The sum of 3 consecutive integers is 38 more than the first integers. Find the 3 integers.

Sagot :

AwesomeHelperidn
Let x be the first integer
Let x + 1 be the second integer
Let x + 2 be the third integer

(x) + (x + 1) + (x + 2) = 38 + (x) + (x + 1)
3x + 1 + 2 = 38 + 2x + 1
3x + 3 = 39 + 2x
3x + 3 - 3 = 39 - 3 + 2x
3x = 36 + 2x
3x - 2x = 36 + 2x - 2x
x = 36 (first integer)

36 + 1 = 37 (second integer)
36 + 2 = 38 (third integer)

(36 + 37 + 38) = 38 + (36 + 37)
(73 + 38) = 38 + (73)
111 = 111

Check:
When 36, 37, and 38 is added to each other the sum is 111, when 36 and 37 is added the sum is 73 and when it is added to 38, it is also 111.
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If it is "first integer" and you just made a mistake in "s" then:

Let x be the first integer
Let x + 1 be the second integer
Let x + 2 be the third integer

(x) + (x + 1) + ( x + 2) = 38 + x
3x + 1 + 2 = 38 + x
3x + 3 = 38 + x
3x + 3 - 3 = 38 - 3 + x
3x = 35 + x
3x - x = 35 + x - x
2x = 35
2x(1/2) = 35(1/2)
x = 17.5 or 17 1/2 (first integer)

17 1/2 + 1 = 18 1/2 or 18.5 (second integer)
17 1/2 + 2 = 19 1/2 or 19.5 (third integer)

17.5 + 18.5 + 19.5 = 38 + 17.5
36 + 19.5 = 55.5
55.5 = 55.5

Check:
17.5, 18.5, 19.5 when added is 55.5, when 38 is added to the first integer which is 17.5 it will also be 55.5.

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