SOLUTION: 2. 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 prod

Question 84572: 2. 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. 36,63 b. 81, 18 c. 27, 72 d. 45, 54 e. none

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
2.Let 10a+b=one of the two numbers
Now we are told that a+b=9 and b+a=9----eq1
Now we are also told that a^2-ab+b^2=21 and b^2-ab+a^2=21-----eq2
substitute a=(9-b)from eq1 into eq2 and we get:

(9-b)^2-b(9-b)=b^2=21 get rid of parens
81-18b+b^2-9b+b^2+b^2=21 collect like terms
3b^2-27b+81=21 subtract 21 from both sides
3b^2-27b+81-21=21-21 simplify and also divide both sides by 3
b^2-9b+20=0---quadratic in standard form and it can be factored
(b-4)(b-5)=0
b=4
and
b=5
From eq1, when b=4 then a=5
and when b=5 then a=4
Thus, our two numbers are 45 and 54
ck
5+4 and 4+5=9
a^2-ab+b^2=21
4^2-20+5^2=21
41-20=21
21=21
and
5^2-20+4^2=21
41-20=21
21=21
Hope this helps---ptaylor
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