Let's start with the two consecutive numbers. We don't know what are the two numbers that are side by side (consecutive), so we'll use variables but with a twist...instead of x and y, we'll use x for the first number and x+1 for the other number because the other number is 1 more than the value of x. So part of the problem looks like this: (x)+(x+1).
The problem says "the sum of one-third of the first and one-fourth of the second equal 9." One-third of the first would be 1/3(x) and one-fourth of the second would be 1/4(x+1) so if the sum is 9, then the problem is:
1/3(x)+1/4(x+1)=9.
Since we need to solve for x, we must remove the fractions. Right now they have different denominators so we need to make them the same. The least common denominator is 12 so that's the denominator to use, thus...
1/3x becomes 4/12(x) and 1/4(x+1) becomes 3/12(x+1).
Now the equation is 4/12(x)+3/12(x+1)=9
Multiply both sides by 12 so the denominators are gone:
[12•]4/12x+3/12(x+1)=9•12
4x+3(x+1)=108
Using distribution, it's now 4x+3x+3=108. Continuing on with the math...
7x+3=108
7x=105
x=15
We found x! If x=15 the the next number must be 16. Let's check the math: Does 1/3(15)+1/4(16)=9?
5+4=9 and 9=9 so YES! The numbers are 15 and 16
