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

(d.) 45, 54

Step-by-step explanation:

Let the first digit = y

Let the second digit = z

y +z = 9 (1)

y²- yz +z² = 21(2)

From equation (2),

z = 9-y(3)

Substitute equation (3) into (2):

y²- y(9-y) +(9-y)² = 21

y²-9y+y²+y²-18y+81 = 21

3y²-27y+ 81 = 21

3y²-27y+ 81-21= 0

3y²-27y+ 60= 0

y²- 9y +20= 0

(y -5) (y-4) =0

y= 5 or y =4

z = 4 or 5 (substituting into (3))

So the numbers are  54  or 45.

answer: 21$

step-by-step explanation:

$42 / $14 = 3   (ratio 3: 1)

$7 x 3 = $21