tougher problems on simple equations

A fraction is such that if "c" is added to the numerator and "d" to the denominator, the value of the fraction becomes 1/x. If the numerator of the original fraction is doubled and the denominator is increased by "e", the fraction becomes 1/y. Find the original fraction.






Answer:


Let the fraction be "a/b".

So, ( a + c) / ( b + d) = 1/x    ------(1)

or,    x(a + c) = b + d

or,    xa - b = d - xc            ----------(2)


and, 2a/ (b +e) = 1/y      -------------------(3)

or, 2ay = b + e

or, 2ay - b = e   

or, b = 2ay - e --------------(4)

substituting (4) in (2), we get

xa - 2ay + e = d - xc

or, a (x - 2y) = d - xc - e

or, a = (d -xc - e) / (x - 2y)

and b = 2y(d - xc - e) /( x - 2y)   - e

or b = (2yd - 2yxc - ex) / (x - 2y)