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Posted by Subhotosh Khan on October 09, 2002 at 17:22:22:

In Reply to: Algebra 2 posted by Stacy on October 09, 2002 at 16:55:52:

: Can you explain and show me how to do these problems?
: 1. Find three consecutive odd intergers such that 7 times the sum of the first and the third is 120 less than 10 times the opposite of the second?

: 2. Find three consecutive even intergers such that 6 times the sum of the first and the third is 8 less than 14 times the second?

: WE are desperately seeking help. We are homeschooler and I, the mom, have been out of school for a long time! We would appreciate any help you can give us.
: Thanks
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odd numbers are expressed as (2*n+1) where n can be any integer.

Let the first odd integer (in the series) be = 2n + 1

So the next one(odd) would be 2n + 3 and the next one would be 2n + 5

According to the problem"

sum of first and third = (2n + 1) + (2n + 5) = 4n + 6

10*(2n + 3) - (4n + 6) = 120

20n + 30 - 4n - 6 = 120

20n - 4n = 120 - 30 + 6

16n = 96

n = 6

so the numbers are (2*6+1 = )13, 15, 17

check:

13 + 17 = 30

10 * 15 = 150 and

150 - 30 = 120

similarly you can do the second problem. Even integers are expressed as (2n). So the consecutive even integers would be (2n), (2n + 2) & (2n + 4). Carry on from there...

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