Question

If 3 is added to the number and denominator of a fraction and the result subtracted 'by' the original fraction, the difference is 1/15. The numerator of the original fraction is 6 less than the denominator. What is the original fraction?

Answer 1

`9/15`

Explanation:

So if the original fraction's numerator is `6` less than the denominator, we can write the original fraction as `x/(x+6)`.

Then, the second fraction is `3` added to the numerator and denominator. Written like how we wrote the first fraction, it is `(x+3)/(x+6+3)` or `(x+3)/(x+9)`.

We are also told the the difference between the second fraction and the first fraction is `1/15`, so we can write the equation:

`(x+3)/(x+9)-x/(x+6)=1/15`.

Now, we have to find the LCD to be able to work with fractions with different denominators. The LCD in this case would be `(x+6)(x+9)`. If we multiply both sides of the equation by that term, we get:

`(x+6)(x+9)((x+3)/(x+9)-x/(x+6))=1/15(x+6)(x+9)`.

Let's distribute the `(x+6)` first on the left side, to get:

`(x+9)(((x+3)(x+6))/(x+9)) -x)=1/15(x+6)(x+9)`

Then distribute the `(x+9)` on the left side, to get:

`=(x+3)(x+6)-x(x+9)=1/15(x+6)(x+9)`.

Now, we can expand both sides to get:

`x^2+9x+18-x^2-9x=1/15(x^2+15x+54)`.

Now, if we move all the terms to one side and do some simplification, we get:

`0=1/15(x^2+15x+54)-18`.

If we multiply both sides by `15` and combine like terms, we get:

`0=x^2+15x-216`.

Factoring, we get:

`0=(x-9)(x+24)`

Thus, the solutions for x are `9` and `-24`. Using these values for `x`, we get:

`9/15` and `(-24)/-18`. But, the negatives cancel in the second expression, which skews the entire setup (if would have been `(-24)/-18-(-21)/-15` but now its `27/21` and the subtraction does not work.

Quickly checking the first expression, you get `12/18-9/15`, which does equate to `1/15`!