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 - 9". 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 - 9". Replace t and u with 4 and 5 then simplify. 10*4 + 5 = 10*5 + 4 - 9. Simplifying you'd get 45 = 54 - 9 or 45 = 45 which make the statement true. Thus making "t + u = 9 and 10t + u = 10u + t - 9" the answer.

I hope this helps!

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9