Question 118565: 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 _______?
:
"The sum of the two digits = 9"
x + y = 9
or
y = (9-x)
:
"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."
:
Using x^2 and the other number as y
x^2 - xy + y^2 = 21
:
Substitute (9-x) for y:
x^2 - x(9-x) + (9-x)^2 = 21
:
x^2 - 9x + x^2 + (81 - 18x + x^2) = 21; FOILed (9-x)(9-x)
:
Combine like terms on the left:
x^2 + x^2 + x^2 - 9x - 18x + 81 - 21 = 0
:
3x^2 - 27x + 60 = 0; our old friend, the quadratic equation
:
Simplify divide by 3:
x^2 - 9x + 20 = 0
:
Factor it:
(x - 4)(x - 5) = 0
:
The numbers are _______?
:
When x = 4; y = 5
When x = 5; y = 4
:
Check using x = 5
5^2 - 5(4) + 4^2 =
25 - 20 + 16 = 21
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