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Question 135322: A two digit number exceeds the sum of the squares of the digits by 19 and double the product of the digits by 44. Find the sum of the digits of the number.
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! The number [x,y] .. example: x = 2 and y = 3 .. so [2,3] or 23
[x,y] is the number represented by 10x + y
10x + y = x^2 + y^2 + 19
10x + y = 2xy + 44
Now, take the second:
10x + y = 2xy + 44
y - 2xy = 44 - 10x
y(1 - 2x) = 44 - 10x
y = (44 - 10x)/(1 - 2x)
Now, plug:
10x + y = x^2 + y^2 + 19
10x + (44 - 10x)/(1 - 2x) = x^2 + [(44 - 10x)/(1 - 2x)]^2 + 19
Long story short: (7,2) and (3.5,-1.5) .. the decimals are excluded because they don't result with two digits
10x + y
10(7) + 2
72
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