Question

A two digit number exceeds by 19 the sum of the squares of its digits and by 44 the double product of its digits. find the number.

Answer 1

let xand y be the number represented by 10x+y 

10x + y = x^2 + y^2 + 19
10x + y = 2xy + 44
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