Question

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 1

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;

$x^{2}-xy+y^{2}$=21

substituting value of x we have

$(9-y)^{2} -(9-y)y+y^{2} =21$

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