Question 72127: find 4 rational numbers such that the product of the 1st 2nd and 3rd is 54. The 2nd number is 2 less than the 1st number, the 3rd number is 5 less than the 2nd number and the 4th number is 3 less than the 3rd number
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Write an equation for each statement
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find 4 rational numbers such that the product of the 1st 2nd and 3rd is 54.
a*b*c = 54
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The 2nd number is 2 less than the 1st number,
b = (a - 2)
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the 3rd number is 5 less than the 2nd number
c = (b - 5)
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and the 4th number is 3 less than the 3rd number
d = (c - 3)
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Ignore d for the moment, and concentrate on finding a, b, & c
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Get a & c in terms of b:
a - 2 = b
a = (b + 2)
c = (b - 5)
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Substitute in the 1st equation and find b:
a*b*c = 54
(b+2)* b * (b-5) = 54
b(b^2 - 3b -10) = 54
b^3 - 3b^2 - 10b = 54
b^3 - 3b^2 - 10b - 54 = 0
ARe you sure that there are integer solutions to this problem??
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