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!

the ratio of white tiles to blue tiles is 48: 16. the ratio in it's simplest form is 3: 1.

step-by-step explanation:

ratio: take the first number and add a colon next to it, then add the second number after that. make sure the ratio order matches up with the written problem!

simplest form: we can simplify the ratio 48 : 16 by dividing both terms by the greatest common factor (gcf).

the gcf of 48 and 16 is 16.

divide both terms by 16:

48 ÷ 16 = 3

16 ÷ 16 = 1

reducing to simplest form:

48 : 16 = 3 : 1