Question

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

Answer 1

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

Then `7n+(n+1) = 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: `7n+(n+1) = 81`

Subtract `1` from both sides to get:

`8n = 80`

Divide both sides by `8` to get:

`n = 10`

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