The sum of three consecutive odd integers is 76 less then seven times the middle number. find three integers

The sum of three consecutive odd integers is 76 less then seven times the middle number. The three integers are 17, 19 and 21 respeectively

Since each consecutive odd integer is separated by a difference of  2

Let "n" be the first integer

n + 2 be the second integer

n + 4 be the third integer

Given that the sum of three consecutive odd integers is 76 less then seven times the middle number

Which means,

The sum of ( n, n + 2, n + 4) is equal to 76 less than seven times the middle number ( 7(n + 2))

That is,

n + n + 2 + n + 4 = 7(n + 2) - 76

3n + 6 = 7n + 14 - 76

4n = 68

n = 17

So we get:

First integer = n = 17

Second integer = n + 2 = 17 + 2 = 19

Third integer = n + 4 = 17 + 4 = 21

Thus the three consecutive odd integers are 17, 19 and 21 respeectively

your answer would be 10x - 1

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!