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

Answer 1

t+u=9

10t+u=10u+t-9

Step-by-step explanation:

Let the first digit in the original number be t, and the second be u.

So, t+u=9

We can write the first number as 10t+u and the second number as 10u+t

Now, we just need to subtract 9 from the second expression and set them equal

10t+u=10u-9

And, yes

45 and 54