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Question 61402This question is from textbook An Incremental Development
: Find three consecutive odd integers such that three times the sum of the first and third is 18 greater than 4 times the second
This question is from textbook An Incremental Development
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! Find three consecutive odd integers such that three times the sum of the first and third is 18 greater than 4 times the second
Odd numbers are 2 apart from each other:
1,3,5,7...
1,1+2,1+2+2,1+2+2+2
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Let the first number be: x
Then the second will be: x+2
Then the third will be:x+2+2=x+4
"3 times" means 3*
"sum" means +
"is" means =
"18 greater" means +18
"4 times" means 4*
:
3(x+x+4)=4(x+2)+18
3(2x+4)=4x+8+18
6x+12=4x+26
6x-4x+12=4x-4x+26
2x+12=26
2x+12-12=26-12
2x=14
2x/2=14/2
x=7
The first number is: x=7
The second number is: x+2=7+2=9
The third number is: x+4=7+4=11
:
Sanity Check, does 3 times the sum of the first and third number=4 time the second +18?
3(7+11)=4(9)+18
3(18)=36+18
54=54 Looks valid.
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Happy Calculating!!!
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