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. Any help plz?

From the definition of the problem, we have a pair of simultaneous equations:

 

[math]\frac{m+1}{n+1}=\frac{3}{2}[/math] and [math]\frac{m^2-1}{n^2-1}=\frac{21}{8}[/math].

 

We also have to assume that [math]n\neq-1[/math] otherwise it doesn't work.

 

Now by simply cross multiplying and rearranging the equations generally we get:

 

[math]2m-3n=1[/math] which implies [math]m=\frac{1}{2}(1+3n)[/math] from the first equation and [math]21n^2-8m^2=13[/math] from the second.

 

By substution, therefore,

 

[math]3n^2-12n-15=0[/math]. The solutions for this are n = 5 or n = -1. But we stated in the question that n can't be -1, so disregard this answer. So n=5, which means m = 8.

ah ok, i was on the right track afterall. I just forgot to square what i subsiuted in for m, i just multiplied by 8 and then used the quadratic formula on the resulting equation, which gave me the surd.

That would cause problems