Hi Sylvia,

I am going to solve a similar problem.

The denominator of a fraction is 1 more than 3 times the numerator. If the denominator is doubled and the numerator is increased by 2, the value of the resulting fraction is 1/4. Find the original fraction.

You don't know either the numerator or denominator of the fraction so let the numerator be n and the denominator be d. The first sentence says

The denominator of a fraction is 1 more than 3 times the numerator.

Translating that into the language of algebra it is

d = 3n + 1

Hence the fraction can be written

n/d = n/(3n + 1)

Now look at the second sentence. Double the denominator [2(3n + 1)] and add 2 to the numerator [n + 2] and form the resulting fraction [(n + 2)/(6n + 2)]. You are told that the value of this fraction is 1/4 which means that the fraction (n + 2)/(6n + 2) is equivalent to the fraction 1/4. Hence if you know the value of n and form (n + 2)/(6n + 2) there will be some cancellation of factor in the numerator and denominator to obtain the result 1/4. In other words

(n + 2)/(6n + 2) = k/4k = 1/4 for some number k.

Thus n + 2 = k and 6n + 2 = 4k. Substituting the first equation into the second gives

6n + 2 = 4k = 4(n + 2)

or

6n + 2 = 4n + 8

and thus

n = 3 and d = 3n + 1 = 10.

Hence the original fraction is 3/10.

Check:

Is the denominator of a fraction is 1 more than 3 times the numerator?

yes!

Double the denominator and increase the numerator by 2. Is the result 1/4?

yes! (3 + 2)/(2 × 10) = 5/20 = 1/4.

Now try your problem,
Penny