Bryn Posted March 23, 2004 Share Posted March 23, 2004 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? Link to comment Share on other sites More sharing options...
Dave Posted March 23, 2004 Share Posted March 23, 2004 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. Link to comment Share on other sites More sharing options...
Bryn Posted March 23, 2004 Author Share Posted March 23, 2004 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. Link to comment Share on other sites More sharing options...
Dave Posted March 23, 2004 Share Posted March 23, 2004 That would cause problems Link to comment Share on other sites More sharing options...
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