Question

Find three consecutive odd integers such that the sum of the first, twice the second, and three times the third is 70?

I'm supposed to write the equation for this word problem. The answer has even numbers. I don't understand.

Answer 1

See below.

Explanation:

Let's say that the first odd integer is `x`. The second consecutive odd integer would have to be `x+2`. (It would not be `x+1` because that would result in an even integer. The sum of any two odd numbers is even.) The third consecutive odd integer would be `(x+2)+2`, or `x+4`.

The sum of the first, twice the second, and three times the third can be written as

`x + 2(x+2) + 3(x+4)`

This equals `70`. We can now distribute and solve for `x`.

`x + 2(x+2) + 3(x+4) = 70`

`x+ 2x + 4 + 3x + 12 = 70` `->` distribute

`6x + 16 = 70` `->` gather like terms

`6x = 54` `->` subtract `16` from both sides

`x = 9` `->` divide both sides by `6`

Thus, the three consecutive odd integers are `9, 11,` and `13`.