At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

In a 2-digit number, the tens digit is 5 less than the units digit. If you reverse the number, the result is 7 greater than double the original number. Find the original number.

Sagot :

The original number is 38

A 2-digit number can be written as:

N = a*10 + b*1

Where a is the tens digit, and b is the units digit, these two are single-digit numbers.

We know that:

"the tens digit is 5 less than the units digit."

This means that:

a = b - 5

(notice that a must be larger than zero and smaller than 10, from this, we can conclude that b is a number in the range {6, 7, 8, 9})

"If you reverse the number, the result is 7 greater than double the original number"

The reverse number is:

b*10 + a

and this is 7 greater than 2 times the original number, then:

b*10 + a = 7 + 2*(a*10 + b)

Then we found two equations:

a = b - 5

b*10 + a = 7 + 2*(a*10 + b)

Replacing the first equation in the second, we get:

b*10 + (b - 5) = 7 + 2*((b - 5)*10 + b)

Now let's solve that:

b*10 + b - 5 = 7 + 2*(11*b - 50)

11*b - 5 = 7 + 22*b - 100

-5 - 7 + 100 = 22*b - 11*b

88 = 11*b

88/11 = b = 8

Now that we know that b = 8, we can use the equation:

a= b - 5

a = 8 -5 = 3

Then the original number is:

a*10 +  b = 3*10 + 8 = 38

The original number is 38

If you want to read more about this, you can see:

https://brainly.com/question/19902993