Answer:
d.45,54
Step-by-step explanation:
let the two numbers be x and y
x+y=9
y+x=9
let x=9-y
the square of either number minus the product of both digits plus the sqaure of the other digit is expressed as;
[tex]x^{2}-xy+y^{2}[/tex]=21
substituting value of x we have
[tex](9-y)^{2} -(9-y)y+y^{2} =21[/tex]
we have;
81 - 18y + y^2 - 9y + y^2 + y^2 = 21
3y^2 - 27y + 60 = 0, dividing through by 3 we get
y^2 - 9y + 20 = 0 using factorization
(y-5)(y-4) = 0
Two solutions
y = 5, then x = 4
y = 4, then x = 5
:
The two numbers would be 54, and 45