The sum of the digits of a certain two-digit number is 9. when you reverse its digits you increase the number by 63. what is the number?

The given such  number is 18.

Step-by-step explanation:

Here, sum of the two digits is 9

Let us assume the tens digit  = m

So, the unit's digit of the number = 9 - m

Now, the current value of the number = 10 ( Tens Digit ) + 1 ( units digit )

= 10 (m) + 1(9-m)  = 10 m +  9 -  m  = 9 m + 9    (a)

Second Case:

Here after reversing the digits, the ten's digit = 9 - m

The unit's digit = m

So, the new value of the number = 10 (Tens digit) + 1( Unit Digit)

= 10 ( 9- m) + m  = 90 - 10 m + m = 90 - 9 m   (b)

Now, according to the question:

The initial Number value  + 63  = The new number value

or, 9 m + 9 + 63  = 90 - 9 m

or, 18 m = 99 - 72 =  18

or, m = 18/18 = 1

or, m = 1

Hence the tens digit of the number = m  = 1

Also, the ones digit = 9 - 1 = 9 - 1 = 8

So, the given number is 18

the answer to the question is a

step-by-step explanation:

it has the same slope and the same y-intercept.

y + 11 = -7/9(x - 18)

y + 11 =-7/9x + 14

b

y = -7/9x + 3

y-intercept (0, 3)