this needs to be solved using substitution

Question

 The sum of the digits of a two digit number is one third of the number. The units digit is five more than the tens didgit. What is the number? 

Answer 1

A 2-digit number is made up of a units digit and a tens digit.

 

If we let a be the units digit and b be the tens digit then the number can be shown as 

10b + a

 

We are told that the sum of the digits is 1/3 of the number

So a + b = (10b+a)/3

 

We also are told that the units digit is 5 more than the tens digit.

So a = b + 5

 

We can use this 2nd equation, a = b+10 and substitute it into the first equation, a + b = (10b+a)/3 like this

(b+5) + b = [10b + (b+5)]/3

 

Let's combine terms

2b + 5 = (11b + 5)/3

 

Continuing to rearrange terms, we multiply both sides by 3

6b + 15 = 11b + 5

Subtract 6b from both sides

15 = 5b + 5

Subtract 5 from both sides

10 = 5b

Divide both sides by 5

b = 2, the tens digit

 

Since a = b+5, then a = 7, the units digit.

 

So the number is 27.

 

Now let's check.

Is the sum of the two digits equal to 1/3 of the original number.

2 + 7 = 9

9 is 27/3

 

Hope this helps.