SOLUTION: Find two integers whose product is 405 such that one of the integers is three less than twice the other integer

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Question 978389: Find two integers whose product is 405 such that one of the integers is three less than twice the other integer
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!

Let  x  and  y  be our numbers.

Then we have a system of two equations

system%28xy+=+405%2C%0D%0Ax+=+2y-3%29.

Substitute the second equation into the first one,  and you get
%282y-3%29%2Ay = 405.

Simplify it:
2y%5E2+-+3y+-+405+=+0.

Now apply the quadratic formula

y%5B1%5D = %283+%2B+sqrt%289%2B4%2A2%2A405%29%29%2F4%29 = %283+%2B+sqrt%283249%29%29%2F4 = %283+%2B+57%29%2F4 = 15,

y%5B2%5D = %283+-+sqrt%289%2B4%2A2%2A405%29%29%2F4%29 = %283+-+sqrt%283249%29%29%2F4 = %283+-+57%29%2F4 = -13.5.

Since we are asked about the integer solutions,  only  y%5B1%5D=3  satisfies.
It gives  x = 2y+-+3 = 2%2A15-3 = 27.

Answer.  The numbers  15  and  27  are the solution.