SOLUTION: Find 3 consecutive odd integers such that four times the sum of the first and second is 1 less than 7 times the third.
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Question 184348
:
Find 3 consecutive odd integers such that four times the sum of the first and second is 1 less than 7 times the third.
Answer by
solver91311(24713)
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Let the first odd integer be
x
.
Then the second odd integer must be
x
+ 2.
Then the third odd integer must be
x
+ 2 + 2 or
x
+ 4.
The sum of the first and the second: (
x
) + (
x
+ 2) or 2
x
+ 2.
Four times the sum of the first and the second: 4(2
x
+ 2) or 8
x
+ 8.
"is" means equals.
Seven times the third: 7(
x
+ 4) or 7
x
+ 28.
One less than seven times the third: 7
x
+ 28 - 1 or 7
x
+ 27
Putting it all together:
Add -7
x
and -8 to both sides:
So the first odd integer is 19, the second is 2 more than that or 21, and the third is 2 more than that or 23.
Check:
The sum of the first and the second: 19 + 21 = 40.
Four times the sum of the first and the second: 4 X 40 = 160.
Seven times the third: 7 X 23 = 161, less 1 = 160.
Answer checks.
John
