document.write( "Question 28966: 3-2/3x_>1 \n" ); document.write( "
Algebra.Com's Answer #15868 by sdmmadam@yahoo.com(530)\"\" \"About 
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3-2/3x_>1\r
\n" ); document.write( "\n" ); document.write( "3-(2/3)X(1/x) >= 1
\n" ); document.write( "(of course x is NOT ZERO here) (as division by zero is not defined)
\n" ); document.write( "3-1 >= (2/3)X(1/x)
\n" ); document.write( "2 >= (2/3)X(1/x)
\n" ); document.write( "1 >= (1/3x) ---(*)
\n" ); document.write( "Case 1: Let x >0
\n" ); document.write( "Then multiplying (*) by (x) >0
\n" ); document.write( "(multiplying by a positive quantity does not alter the inequality sign)
\n" ); document.write( "(And here greater than remains greater than.)
\n" ); document.write( "(1)X(x) >= (1/3x)X(x)
\n" ); document.write( " x >= 1/3
\n" ); document.write( "We have got the result x >= 1/3 for our consideration of x >0
\n" ); document.write( "Anything equal to 1/3 or to the right of 1/3 is of course positive.
\n" ); document.write( "And as something positive (that is to the right of zero but to the left of 1/3 cannot be >= to 1/3.
\n" ); document.write( "So the verdict is x >= 1/3\r
\n" ); document.write( "\n" ); document.write( "Case 2: Let x <0
\n" ); document.write( "Then multiplying (*) by (x) < 0
\n" ); document.write( "(multiplying by a negative quantity alters the inequality sign)(
\n" ); document.write( "And here greater than becomes less than.)
\n" ); document.write( "(1)X(x) =< (1/3x)X(x)
\n" ); document.write( " x =< 1/3
\n" ); document.write( "We have got the result x =< 1/3 for our consideration of x < 0
\n" ); document.write( "Anything equal to 1/3 is not negative and anything to the left of 1/3 need not be negative.
\n" ); document.write( "And as something negative is always less than 1/3, the verdict is x < 0
\n" ); document.write( "Answer: Either x >=1/3 or x<0\r
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