# How do you find the two consecutive integers whose sum is 223?

May 22, 2018

See a solution process below:

#### Explanation:

First, let's cal the first integer we are looking for: $n$

Then, because we are looking for consecutive integers the second integer we are looking for can be written as: $n + 1$

We know these two integers sum to 223. Therefore, we can write this equation and solve for $n$:

$n + \left(n + 1\right) = 223$

$n + n + 1 = 223$

$1 n + 1 n + 1 = 223$

$\left(1 + 1\right) n + 1 = 223$

$2 n + 1 = 223$

$2 n + 1 - \textcolor{red}{1} = 223 - \textcolor{red}{1}$

$2 n + 0 = 222$

$2 n = 222$

$\frac{2 n}{\textcolor{red}{2}} = \frac{222}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} n}{\cancel{\textcolor{red}{2}}} = 111$

$n = 111$

• The First integer is: $111$

• The Second integer is: $111 + 1 = 112$

Solution Check:

$111 + 112 = 223$