Chapter 4.2, Problem 12P

To determine

To find: four consecutive integers such that the sum of the two greatest is 17 less than twice the

sum of the two smallest

Answer to Problem 12P

The required answer is: consecutive integers are 10, 11, 12 and 13

Explanation of Solution

Calculation:

Let the first integer be x

Let the second integer be x+1

Let the third integer be x+2

Let the fourth integer be x+3

  $(x+2)+(x+3)=2(x+(x+1))-17$

(Sum of two greatest integers is 17 less than twice the sum of the two smallest)

  $x+2+x+3=2x+2x+2-17$

(Removing the bracket)

  $\begin{array}{l}2x+5=4x-15\\ 2x+5-2x=4x-15-2x\end{array}$

(Subtract 2x from both sides)

  $\begin{array}{l}5=2x-15\\ 5+15=2x-15+15\end{array}$

(Add 15 to both sides)

  $\begin{array}{l}20=2x\\ \frac{20}{2}=\frac{2x}{2}\end{array}$

(Divide 2 on both sides)

  $10=x$

(Simplify)

First integer:

  $x=10$

Second integer:

  $\begin{array}{l}x+1=10+1\\ x+1=11\end{array}$

Third integer:

  $\begin{array}{l}x+2=10+2\\ x+2=12\end{array}$

Fourth integer:

  $\begin{array}{l}x+3=10+3\\ x+3=13\end{array}$

Therefore the consecutive integers are 10, 11, 12 and 13.