find three consecutive odd integers such that six times the second decreased by twice the first is equal to twenty more than the sum

Question

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Answer 1

find three consecutive odd integers such that six times the second decreased by twice the first is equal to twenty more than the sum

The important part about a question like this is to set one of the integers to be equal to a variable in order to solve the problem. For example, let's say that the smallest integer of the three is x. We don't actually even need to set variables for the other integers because the question tells us that they are consecutive odd integers, which means the following:

x = smallest integer

x + 2 = middle integer

x + 4 = largest integer

Why does that make sense? For example, let's say the smallest integer is 5. That means the consecutive odd integers after 5 would be 7 and 9, and 7 is the same thing as saying 5 + 2 and 9 is the same as saying 5+4.

Now that we've figured out how to write all the integers in a way that only uses the variable x, let's plug in all those values into the word problem.

"Six times the second decreased by twice the first is equal to twenty more than the sum" can be rewritten as:

6 * second - 2 * first = first + second + third + 20

6 * (x + 2) - 2 * x = x + (x + 2) + (x + 4) + 20

6x + 12 - 2x = 3x + 26

4x + 12 = 3x + 26

x + 12 = 26

x = 14

That means that the smallest integer is 14, so the other integers are 16 and 18. However, the question asked for ODD consecutive integers, so in this there is no answer.

Answer: No possible integers, only possible if even (14, 16, 18)

Answer 2

6s- 2f= 20+f+s+t

where s = 2nd,=x+2 f=1st =x, t =3rd = x+4

6(x+2)-2(x)= 20+ x+x+2+x+4

6x+12-2x= 20+3x+6

4x+12=26+3x

x=14

nope, 14, 16, 18 are not odd

but these are 3 consecutive even numbers

96-28= 20+48

68=68

or maybe you meant sum of f+s, sum of just the first and second, then

6s-2f=20+f+s

6x+12-2x= 20+x+x+2

2x=10

x=5

5, 7, 9

6(7)-2(5)=20+5+7

42-10= 32

32=32