document.write( "Question 616808: how do i solve this problem (x-1)/(x+2)>0 \n" ); document.write( "
Algebra.Com's Answer #387917 by jsmallt9(3758)\"\" \"About 
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\n" ); document.write( "The solution to this is based on some simple logic:
  • This inequality, in its essence, says that a fraction is positive.
  • There are only two situations that make a fraction positive:
    • The numerator and denominator are both positive; or
    • The numerator and denominator are both negative.
All we have to do is translate this logic into some inequalities and solve.

\n" ); document.write( "\"the numerator and denominator are both positive\" translates into:
\n" ); document.write( "((x-1) > 0 and (x+2) > 0)
\n" ); document.write( "\"the numerator and denominator are both negative\" translates into:
\n" ); document.write( "((x-1) < 0 and (x+2) < 0)
\n" ); document.write( "\"the numerator and denominator are both positive or the numerator and denominator are both negative\" translates into:
\n" ); document.write( "((x-1) > 0 and (x+2) > 0) or ((x-1) < 0 and (x+2) < 0)

\n" ); document.write( "Now we solve.
\n" ); document.write( "(x > 1 and x > -2) or (x < 1 and x < -2)

\n" ); document.write( "(x > 1 and x > -2) will simplify. The only way x can be greater than 1 and -2 at the same time is if x > 1. So we can replace (x > 1 and x > -2) with just (x > 1).

\n" ); document.write( "Using similar logic on (x < 1 and x < -2) we find that we can replace it with just (x < -2)

\n" ); document.write( "So our \"reduced\" solution is:
\n" ); document.write( "(x > 1) or (x < -2)
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