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

Question 199854: 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 ankor@dixie-net.com(22740) (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?
:
A 2 digit number:
a = the 10s digit of the 1st number
b = units of the 1st number
:
"sum of both digits of either of two two-digit numbers, in whatever order the
digits are written, is 9"
a + b = 9
or
a = (9-b)
:
"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."
a^2 - ab + b^2 = 21
Substitute (9-b) for a in the above
(9-b)^2 - b(9-b) + b^2 = 21
81 - 18b + b^2 - 9b + b^2 + b^2 = 21
Arrange as a quadratic equation
3b^2 - 27b + 81 - 21 = 0
3b^2 - 27b + 60 = 0
Simplify, divide by 3
b^2 - 9b + 20 = 0
Factor
(b-5)(b-4) = 0
Two solutions
b = 5, then a = 4
b = 4, then a = 5
:
The two numbers would be 54, and 45

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