The sum of the two digits of a number is 16. the number formed by reversing the digits is 18 more than the original
number. determine the original number.
let t = the tens digit, u = the units digit, and u + t = 16. which of the following equations would complete the system?

10u+t=18+10t+u

(If this doesn't match one of the equations, please post your options so I can put it in the form of one of your options.)

Step-by-step explanation:

Let there be a number [tu] where t is the tens and u is ones digit.

t+u=16.

Now we reverse the number [ut] is 8 more than [tu].

[ut]=18+[tu]

So [ut]=10u+t because u is the tens  and t is the ones digit

and [tu]=10t+u because t is the tens and u is the ones digit.

The equation you are looking for is

10u+t=18+10t+u.

Let's solve it just for fun:

Subtract u on both sids:

9u+t=18+10t

Subtract 10t on both sides:

9u-9t=18

Divide both sides by 9:

u-t=2

The system is:

u-t=2

u+t=16

add!

2u  =18

u=9

If u=9 and u+t=16, then t=7.

To the original number is 79.

The new number is 97 (is this 18 more than the other; yep).

just needed some points but try socratic