Here we have a simple algebra problem, we will see that the original number was 46.
We know that we start with a 2 digit number, it can be written as:
a*10 + b
Where a and b are decimal numbers. (from 0 to 9).
If we add a 3 "in front of it", we are adding:
3*100 + a*10 + b
And this is 24 more than 7 times the original number, so we have:
3*100 + a*10 + b = 24 + 7*(a*10 + b)
Now we can solve that:
300 + a*10 + b = 24 + a*70 + b*7
300 - 24 + a*10 - a*70 + b - b*7 = 0
276 - a*60 - b*6 = 0
276 = a*60 + b*6
This part is tricky, here you need to remember that a and b can only be integers from 0 to 9, so you can try with different integer values for b and see if a is also an integer for that value, for example taking b = 6 we get:
276 = a*60 + 6*6 = a*60 + 36
276 - 36 = a*60
240/60 = a = 4
So if b = 6, then a = 4, so both are integers in the wanted range.
From this, we can conclude that the original number is 46.
If you want to learn more about algebra, you can read:
https://brainly.com/question/4344214
