Question

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?

Answer 1

`v >= 3` and `v+1 >= 4`

Explanation:

First let's get or define our variables. We can call the first variable `v`. Then because the problem states they are two consecutive integers which means the second integer is one more the first integer the second integer can be called `v + 1`.

"Nine times the smaller" can be written as `9v` and "5 times the larger" can be written as `5(v + 1)`.

"is at most" means we have an inequality and specifically a `<=` or less than or equal to inequality.

So, to write the inequality for the entire problem we have:

`v + (v + 1) <= 9v - 5(v + 1)`

Expanding the terms in parenthesis and then grouping like terms on each side of the inequality gives:

`v + v + 1 <= 9v - 5v - 5`

`2v + 1 <= 4v - 5`

Next we solve for `v` while keeping the inequality balanced:

`2v + 1 - 2v + 5 <= 4v - 5 - 2v + 5`

`6 <= 2v`

`6/2 <= (2v)/2`

`3 <= v`

To "flip" the inequality so the `v` is on the left side we have to "flip " the inequality so `<=` becomes `>=` and the inequality can be written as:

`v >= 3`