I think I could possibly be mixing stuff up not sure.... Solve: (x + 1)/(x - 2) > 3 x < 2 or x > 5/2
J josh123 Junior Member Joined Oct 5, 2005 Messages 52 Dec 1, 2005 #1 I think I could possibly be mixing stuff up not sure.... Solve: (x + 1)/(x - 2) > 3 x < 2 or x > 5/2
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Dec 1, 2005 #2 \(\displaystyle \L\frac{{x + 1}}{{x - 2}} - 3 > 0\) \(\displaystyle \L\frac{{\left( {x + 1} \right) - 3(x - 2)}}{{x - 2}} > 0\) \(\displaystyle \L\frac{{ - 2x + 7}}{{x - 2}} > 0\) Now solve it.
\(\displaystyle \L\frac{{x + 1}}{{x - 2}} - 3 > 0\) \(\displaystyle \L\frac{{\left( {x + 1} \right) - 3(x - 2)}}{{x - 2}} > 0\) \(\displaystyle \L\frac{{ - 2x + 7}}{{x - 2}} > 0\) Now solve it.
J josh123 Junior Member Joined Oct 5, 2005 Messages 52 Dec 1, 2005 #3 2 < x < 7/2 ? sorry dont know where i got 5 now that i look at it
