SOLUTION: What are five consecutive odd integers from least to greatest in which the sum of the first and the fifth is one less than three times the fourth?

Question 499204: What are five consecutive odd integers from least to greatest in which the sum of the first and the fifth is one less than three times the fourth?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Five consecutive odd integers could be called: a, b, c, d, and e. That would not be a good choice. Instead, show them as:
.
x
x+2
x+4
x+6
x+8
.
As long as 'x' is odd, then the next consecutive integer will be +2. Etc.
.
The sum of the first and fifth;
x + x + 8 = 2x +8
.
One less than 3 times the fourth:
3*(x+6) -1
.
These two expressions are equal:
2x+8 = 3*(x+6) -1
2x+8 = 3x + 18 -1
.
Subtract 8 from both sides
2x = 3x + 18 -1 - 8
.
Subtract 3x from both sides
2x -3x = 18 -1 -8
.
Collect terms
-x = 9
.
Multiply both sides by -1
x = -9
.
Therefore,
x = -9
x+2 = -7
x+4 = -5
x+6 = -3
x+8 = -1
.
They're consecutive odd integers.
.
What is the sum of the first and fifth?
-9 + (-1) = -10
.
What is 3 times the fourth element -1?
3(-3) -1 = -9 -1 = -10
.
These all match up with the problem statement and the required answer.
.
Done
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