Question

if we write a 3 in front of a two-digit number, the new three digit number will be 24 more than 7 tines the origional two digit number. What was that number

Answer 1

Here we have a simple algebra problem, we will see that the original number was 46.

How to find the equation we need to solve:

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:

brainly.com/question/4344214