Posted by MathBard on October 09, 2002 at 19:24:07:
In Reply to: Hints for #1 posted by Subhotosh Khan on October 09, 2002 at 17:22:22:
: : 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
: 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
: 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...
i recognize that your are a brilliant Mathematician, but you misread her problem.
You didnt notice that the problem said... the opposite of 10 time the OPPOSITE of the second.
here is how i did it.
let x equal first odd integer.
X+2 second odd integer
7(x+x+4) is seven time the sum of first and second whis is 120 less than...
7(x+x=4) + 120=
ten times the Opposite of the second
-7 -5 -3 are your consecutive odd ints
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