# How do you find two consecutive integers such that seven times the first integer plus the second is 81?

Jun 15, 2015

Let the integers be $n$ and $n + 1$.

Then $7 n + \left(n + 1\right) = 81$. Hence $n = 10$ and $n + 1 = 11$

#### Explanation:

Let $n$ stand for the first integer.

Then the second integer is $n + 1$.

We are given: $7 n + \left(n + 1\right) = 81$

Subtract $1$ from both sides to get:

$8 n = 80$

Divide both sides by $8$ to get:

$n = 10$

So the two integers are $n = 10$ and $n + 1 = 11$