# How do you find four consecutive integers such that the sum of the two largest subtracted from twice the sum of the two smallest is 15?

Jun 18, 2015

Substitute $n$, $n + 1$, $n + 2$ and $n + 3$ into the description to get an equation in $n$ and solve it to find $n = 9$.

Hence the integers are: $9$, $10$, $11$ and $12$.

#### Explanation:

Let the four integers be $n$, $n + 1$, $n + 2$ and $n + 3$

The sum of the two largest integers is:

$\left(n + 2\right) + \left(n + 3\right) = 2 n + 5$

Twice the sum of the two smallest integers is:

$2 \left(n + \left(n + 1\right)\right) = 2 \left(2 n + 1\right) = 4 n + 2$

We are given:

$15 = 2 \left(n + \left(n + 1\right)\right) - \left(\left(n + 2\right) + \left(n + 3\right)\right)$

$= \left(4 n + 2\right) - \left(2 n + 5\right) = 2 n - 3$

Add $3$ to both ends to get: $2 n = 18$

Divide both sides by $2$ to get $n = 9$

So the four integers are: $9$, $10$, $11$, $12$