# One integer is nine more than two times another integer. If the product of the integers is 18, how do you find the two integers?

May 19, 2018

Solutions integers: $\textcolor{b l u e}{- 3 , - 6}$

#### Explanation:

Let the integers be represented by $a$ and $b$.
We are told:
[1]$\textcolor{w h i t e}{\text{XXX}} a = 2 b + 9$ (One integer is nine more than two time the other integer)
and
[2]$\textcolor{w h i t e}{\text{XXX}} a \times b = 18$ (The product of the integers is 18)

Based on [1], we know we can substitute $\left(2 b + 9\right)$ for $a$ in [2];
giving
[3]$\textcolor{w h i t e}{\text{XXX}} \left(2 b + 9\right) \times b = 18$

Simplifying with the target of writing this as a standard form quadratic:
[5]$\textcolor{w h i t e}{\text{XXX}} 2 {b}^{2} + 9 b = 18$

[6]$\textcolor{w h i t e}{\text{XXX}} 2 {b}^{2} + 9 b - 18 = 0$

You could use the quadratic formula to solve for $b$ or recognize the factoring:
[7]$\textcolor{w h i t e}{\text{XXX}} \left(2 b - 3\right) \left(b + 6\right) = 0$
giving solutions:
$\textcolor{w h i t e}{\text{XXX}} b = \frac{3}{2}$ which is not permitted since we are told the values are integers.
or
$\textcolor{w h i t e}{\text{XXX}} b = - 6$

If $b = - 6$ then based on [1]
$\textcolor{w h i t e}{\text{XXX}} a = - 3$