Question

How do you find three consecutive odd integers such that he sum of the first and twice the second is 6 more than the third?

Answer 1

See process for finding the solution to this problem below:

Explanation:

First, let's define the three consecutive odd numbers.

The first, we can call `x`.

Because they are odd numbers we know they are every other number from `x` so we need to add `2`

The second and third number will be `x + 2` and `x + 4`

"twice the second" is the same as: `2(x + 2)`

"the sum of the first and twice the second is then: `x + 2(x + 2)`

This sum is "6 more than the third" or `(x + 4) + 6`

Putting this into an equation and solving for `x`:

`x + 2(x + 2) = (x + 4) + 6`

`x + 2x + 4 = x + 10`

`3x + 4 = x + 10`

`3x + 4 - color(red)(x) - color(blue)(4) = x + 10 - color(red)(x) - color(blue)(4)`

`3x - color(red)(x) + 4 - color(blue)(4) = x - color(red)(x) + 10 - color(blue)(4)`

`2x + 0 = 0 + 6`

`2x = 6`

`(2x)/color(red)(2) = 6/color(red)(2)`

`(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 3`

`x = 3`

The first number is 3 and therefore the next two consecutive odd numbers are 5 and 7