Chapter 4.2, Problem 13P

To determine

To find: four consecutive even integers such that the fourth is the sum of the first and second.

Answer to Problem 13P

The required answer is: consecutive even integers are 4, 6, 8 and 10

Explanation of Solution

Calculation:

Let the first even integer be x

Let the second even integer be x+2

Let the third even integer be x+4

Let the fourth even integer be x+6

  $(x+6)=(x)+(x+2)$

(Fourth is the sum of first and second)

  $\begin{array}{l}x+6=2x+2\\ x+6-x=2x+2-x\end{array}$

(Subtract x from both sides)

  $\begin{array}{l}6=x+2\\ 6-2=x+2-2\end{array}$

(Subtract 2 from both sides)

  $4=x$

(Simplify)

First even integer:

  $x=4$

Second even integer:

  $\begin{array}{l}x+2=4+2\\ x+2=6\end{array}$

Third even integer:

  $\begin{array}{l}x+4=4+4\\ x+4=8\end{array}$

Fourth even integer:

  $\begin{array}{l}x+6=4+6\\ x+6=10\end{array}$

Therefore the consecutive even integers are 4, 6, 8 and 10.