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 t
han 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
1 answer:
Answer:
t+u=9
10t+u=10u+t-9
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
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