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?

1 Answer

Answer:

#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#!