Expert: Paul Klarreich - 11/3/2008

Question
Hello,
I have this question and I think I have come up with the right formula, however, I am having trouble solving it.
Can you help?

Find 4 consecutive integers such that 39 less than 4 times the sum of the 2 middle integers is the same as half the sum of the 3 largest integers decreased by the smallest integer.
Write an equation and solve.

Here's what I came up with, but I know that something is wrong.

4(x+1+x+2)-39=x+1+x+2+x+3-x/2
4(2x+3)-39
8x+12-39
8x-27=x+6
7x-27=6
7x=33


Answer
Questioner:   Debbi
Category:  Advanced Math
Private:  No
 
Subject:  Algebra
Question:  Hello,
I have this question and I think I have come up with the right formula, however, I am having trouble solving it.
Can you help?

Find 4 consecutive integers such that 39 less than 4 times the sum of the 2 middle integers is the same as half the sum of the 3 largest integers decreased by the smallest integer.
Write an equation and solve.

Here's what I came up with, but I know that something is wrong.

4(x+1+x+2)-39=x+1+x+2+x+3-x/2
4(2x+3)-39
8x+12-39
8x-27=x+6
7x-27=6
7x=33

Hi,Debby,

>> 7x = 33, DOES NOT GIVE AN INTEGER SOLUTION, AS YOU NOTICED.

Let  x, x+1, x+2, x+3   represent the integers:

Then on the left side:

The sum of the two middle integers is  x+1 + x+2 = 2x + 3
Four times that sum = 8x + 12
39 less than that is  8x + 12 - 39 = 8x - 27

For the right side, the interpretation of:

"half the sum of the 3 largest integers decreased by the smallest integer"

is definitely:

"half (the sum of the 3 largest integers) decreased by the smallest integer"

>> because "the sum of" applies to the 3 largest.., not to the "decreased by"

"half (x + 1 + x + 2 + x + 3) decreased by the smallest integer"
"half (x + 1 + x + 2 + x + 3) - x"
"half (3x + 6) - x"
"3x/2 + 3 - x"
"x/2 + 3 "

Let's try that:

8x - 27 = x/2 + 3

Clear fractions:

16x - 54 = x + 6
15x = 60
x = 4, etc.