SOLUTION: The sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9. THe square of either of the digits of either number, minus the product o

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Question 204317: The sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9. THe square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21. The numbers are?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
The sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9. THe square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21. The numbers are?

It's difficult to read that since it talks about two
numbers.   So let's reword it for just ONE of the two 
numbers:

The sum of the digits is 9. The square of the tens 
digits minus the product of both digits, plus the 
square of the one's digit is the number 21.

Let the tens digit be t and ones (or units) digit be u.

>>...The sum of the digits is 9...<<



>>...The square of the tens digits minus the product 
of both digits, plus the square of the one's digit is 
the number 21...<<



So we have the system of equations:



Solve the first equation for u:



Substitute in the other equation:













Divide through by 3





; 
  ;    

So there are two possibilites for t

First possibility:
Substitute  into 





So one solution is 54 

Second possibility:
Substitute  into 





So the other solution is 45.

So these are the two numbers, 54 and 45  

Edwin

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