The mayor gets elected every four years. Treasurer gets elected every six years. Both were elected this year. How many years before the election happens in the same year.

Step-by-step explanation:

The mayor gets selected once in every 4 years.

So, the number of years the mayor elections takes place

=  1, 4 ,8, 12, 16, 20, 24, 28,

Now, the Treasurer gets selected once in every 6 years.

So, the number of years the Treasurer elections takes place

= 1,  6, 12, 18, 24, 30

Now,as we can see from both statements, the year in which both elections happen are the common factors of 4 and 6.

The first common factor of 4 and 6 is 12.

So,in twelfth year both elections happen in the same year again.

Hence, there are 11 years before the elections happen in the same year.

4x and 6y

im not sure though

only 10/60 isn't equivalent

hope this : )