Can somebody tell me how they got 45,54 for this question?

Question

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 _______?

Answer 1

x+y = 9

x^2 - xy + y^2 = 21

Now, in the first equation y = 9 - x

So by substitution:

x^2 - x(9-x) + (9 - x)^2 = 21

Simplifying:

x^2 - 9x + x^2 + 81 - 18x + x^2 = 21

3x^2 - 27x + 60 = 0

3(x^2 - 9x + 20) = 0

3(x - 4) (x - 5) = 0

x= 4 or x = 5

If x = 4, then y = 9-4 = 5, and if x= 5, then y = 4, so the possible answers are 45 and 54.

Answer 2

Let the first number 10t +u , where t= tens digit and u= units digit
Let the 2nd number be 10t1 +u1
t+u=9
t1+u1=9
t^2-tu +u^2=21
t1^2-t1u1 +u1^2=21
Now substiture u = 9-t into equation 3 and get:
t^2-t(9-t) +(9-t)^2 =21
t^2 -9t +t^2 + 81 -18t +t^2=21
3t^2 -27t +60=0
t^2-9t+30 = 0
(t-5)(t-6)=0
so t= 5 or t= 6
so u = 4 or 3
so 10t+u =54 or 63
The number 63 does not satisfy the conditions so the number must be 54.
Since the order of the digits is irrelevant the other number is 45:

Answer 3

rewrite the question in numerical
x + y = 9
x^2 - xy + y^2 = 21
try to sub in the number
x = 9 - y
then plug in to the second equation
(9-y)^2 - y(9-y) + y^2 = 21
81 - 18y + y^2 -9y + y^2 + y^2 = 21
(81 - 27y + 3y^2 = 21)/3
(27 - 9y + y^2 = 7) minus 7 in both sides
20 - 9y + y^2 = 0
(4-y)(5-y) = 0
so y = 4 or y = 5
when you know y, then you plug in the real number to y.
x + 4 = 9if y is 4, x must be 5
x + 5 = 9if y is 5, x must be 4

Answer 4

let a & b be the 2 digits

a + b = 9
a^2 - ab + b^2 = 21

(a + b)^2 = 9^2
a^2 + 2ab + b^2 = 81

a^2 + 2ab + b^2 = 81
-(a^2 - ab + b^2 = 21)
3ab = 60
ab = 20

now remember a + b = 9
let b = 9 - a
a(9- a) = 20

-a^2 +9a - 20 = 0
a^2 - 9a + 20 = 0
(a - 5)(a - 4) = 0
a = 5 or 4
plug it back into a + b = 9
then b can also be 5 or 4

Answer 5

x+y=9
x^2-(xy)+y^2=21
(x+y)^2=81
x^2+2xy+y^2=81
x^2-xy+y^2+3xy=81
21+3xy=81
3xy=60
xy=20
the only factors of 20 satisfying the conditions are4 and 5
so the nos are 45 and 54