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

The answer would be "t + u = 9 and 10t + u = 10u + t". This can be found if you replaced t with 4 and u with 5. Adding the 2 digits would give you 9, and using the formula "10t + u = 10u + t". Replace t and u with 4 and 5 then simplify. 10*4 + 5 = 10*5 + 4. Simplifying you'd get 45 = 54 which gives you 2 digits that equal 9 but when reversed are 9 more than the original number.

The equations described in this problem are:

t + u = 9

10t + u = 10u + t - 9

When the two equations are substituted and solved, t = 4 and u = 5.

The original number described in the word problem is 45.