Chapter 4.2, Problem 11P

To determine

To find: three consecutive odd integers such that twice the smallest is 3 more than the greatest integer

Answer to Problem 11P

The required answer is: consecutive odd integers are -7, -5 and -3

Explanation of Solution

Calculation:

Let the first odd integer be x

Let the second odd integer be x+2

Let the third odd integer be x+4

  $2x=(x+4)+3$

(Twice the smallest is 3 more than the greatest integer)

  $\begin{array}{l}2x=x+7\\ 2x-x=x-7-x\end{array}$

(Subtract x from both sides)

  $x=-7$

(Simplify)

First integer:

  $x=-7$

Second integer:

  $\begin{array}{l}x+2=-7+2\\ x+2=-5\end{array}$

Third integer:

  $\begin{array}{l}x+4=-7+4\\ x+4=-3\end{array}$

Therefore the consecutive odd integers are -7, -5 and -3.