paul jones

Posted on Tuesday, 23 March, 2004 - 08:00 pm:

Was browing online and saw this could do with the matematical answer please:

When 1 is added to the numerator and the denominator of the fraction m/n the new fraction is 3/2. When one is subtracted from the numerator and denominator of the fraction m^2/n^2 the new fraction is 21/8. Find the possible values of m and n.

The values i get for n and m are surds, but the answer is apparently m=8 and n=5, which do work. I wont post my working cos i'm pretty sure i'm totally off track.

Phil Freeman

Posted on Tuesday, 23 March, 2004 - 08:39 pm:

We have (m+1)/(n+1)=3/2
and (m^2-1)/(n^2-1)=21/8
The first gives 2m-3n=1
The second can be factorised on top and bottom as the difference of 2 squares. Dividing by the first equation gives (m-1)/(n-1)=7/4
which gives
4m-7n=-3
We then have 2 simultaneous equations in 2 unknowns, so we can find m and n.

-Phil.

Tristan Marshall

Posted on Tuesday, 23 March, 2004 - 08:44 pm:

The first part tells us that

(m + 1) / (n + 1) = 3 / 2

, which we can re-arrange to get:

$2 m = 3 n + 1$

The second part tells us that $(m^{2} - 1) / (n^{2} - 1) = 21 / 8$ . We can re-arrange this using `difference of two squares' and the first part, thus:

$\frac{21}{8} = \frac{m^{2} - 1}{n^{2} - 1} = \frac{m + 1}{n + 1} \cdot \frac{m - 1}{n - 1} = \frac{3}{2} \cdot \frac{m - 1}{n - 1}$

We can re-arrange this last to obtain:

$4 m = 7 n - 3$ .

Hence we now have two simultaneous equations for $m$ and $n$ , which give the solution $m = 8$ and $n = 5$ .

paul jones

Posted on Tuesday, 23 March, 2004 - 09:15 pm:

Thank you you guys i now see where it was coming from much appreciated.