SOLUTION: Find two integers whose product is 408 such that one of the integers is five less than seven times the other integer.

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Question 409556: Find two integers whose product is 408 such that one of the integers is five less than seven times the other integer.
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi
one of the integers is five less than seven times the other integer.
Let x and (7x-5)represent the intergers
Question states***
x(7x-5) = 408
7x^2 - 5x -408 = 0
Solving for x



x = 112/14 = 8 and x = -102/14 (extraneous solution as is a non-integer)
Integers are 8 and 51 (7*8-5)
CHECKING our Answer***
8*51 =408
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