SOLUTION: find three consecutive even integers such that five-half of the second minus three half of the first is three-fourths of the third integers.

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Question 1043061: find three consecutive even integers such that five-half of the second minus three half of the first is three-fourths
of the third integers.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
2n is the first of these consecutive even integers, and n is any integer.

%285%2F2%29%282n%2B2%29-%283%2F2%29%282m%29=%283%2F4%29%282n%2B4%29

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

find three consecutive even integers such that five-half of the second minus three half of the first is three-fourths
of the third integers.
Let the smallest be S

Then the 2nd and 3rd are: S + 2, and S + 4, respectively

We then get: %285%2F2%29%28S+%2B+2%29+-+%283%2F2%29S+=+%283%2F4%29%28S+%2B+4%29

5%28S+%2B+2%29%2F2+-+3S%2F2+=+3%28S+%2B+4%29%2F4

2(5)(S + 2) - 2(3S) = 3(S + 4) ------- Multiplying by LCD, 4

10S + 20 - 6S = 3S + 12

4S + 20 = 3S + 12

4S - 3S = 12 - 20

S, or smallest integer = highlight_green%28-+8%29

Integer 2 = - 8 + 2, or  highlight_green%28-+6%29

Integer 3 = - 8 + 4, or  highlight_green%28-+4%29

That's all!