# 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?

Nov 11, 2016

$v \ge 3$ and $v + 1 \ge 4$

#### 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 $9 v$ and "5 times the larger" can be written as $5 \left(v + 1\right)$.

"is at most" means we have an inequality and specifically a $\le$ or less than or equal to inequality.

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

$v + \left(v + 1\right) \le 9 v - 5 \left(v + 1\right)$

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

$v + v + 1 \le 9 v - 5 v - 5$

$2 v + 1 \le 4 v - 5$

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

$2 v + 1 - 2 v + 5 \le 4 v - 5 - 2 v + 5$

$6 \le 2 v$

$\frac{6}{2} \le \frac{2 v}{2}$

$3 \le v$

To "flip" the inequality so the $v$ is on the left side we have to "flip " the inequality so $\le$ becomes $\ge$ and the inequality can be written as:

$v \ge 3$