Does a two-digit number exist such that the digits sum to 9 and when the digits are reversed the resulting number is 9 greater than the original number? Identify the system of equations that models the given scenario.

t + u = 9
10t + u = 10u + t – 9

t + u = 9
10t + u = 10u + t

t + u = 9
tu = ut + 9

Question

Does a two-digit number exist such that the digits sum to 9 and when the digits are reversed the resulting number is 9 greater than the original number? Identify the system of equations that models the given scenario.

t + u = 9
10t + u = 10u + t – 9

t + u = 9
10t + u = 10u + t

t + u = 9
tu = ut + 9

mathstudent55 · Answer

t = tens digit
u = units digit

mathstudent55 · Answer

The original number is
10t + u
Ok so far?

macaylam · Answer

yeah

mathstudent55 · Answer

The original number is actually 10t + u.
The sum of its digits is simply t + u
Ok?

mathstudent55 · Answer

The sum of the digits of the original number is 9, so we have
t + u = 9
That is one equation.
Now we need the second equation.

mathstudent55 · Answer

The original number is 10t + u
The new has the digits reversed, so the new number is 10u + t

mathstudent55 · Answer

The new number is 9 greater than the original number.
That means
10u + t is 9 greater than 10t + u or
10u + t = 10t + u + 9

mathstudent55 · Answer

Subtract 9 from both sides of the last equation:

10u + t - 9 = 10t + u

Switch sides:

10t + u = 10u + t - 9

This is our second equation.

The system of equations is:

t + u = 9
10t + u = 10u + t - 9

Which option is that?

macaylam · Answer

A

mathstudent55 · Answer

Correct.

macaylam · Answer

thanks for helping (:

mathstudent55 · Answer

You're welcome.