document.write( "Question 308695: If xy > 1 and z < 0, which of the following statements must be true?
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\n" ); document.write( "I. x > z
\n" ); document.write( "II. xyz < -1
\n" ); document.write( "III. xy/z < 1/z
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Algebra.Com's Answer #220927 by jim_thompson5910(35256) $\"\"$ $\"About$

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I. False. For example, let x = -10 and z = -1. Clearly $\"x%3Ez\"$ is false. If we let $\"y=-5\"$ , then $\"xy=-10%28-5%29=50%3E1\"$ showing that $\"xy%3E1\"$ is true.\r
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\n" ); document.write( "\n" ); document.write( "II. False, this is only true if $\"xyz%3Cz\"$ . So $\"xyz+%3C+-1\"$ is on the right track, but it is false. For example, if $\"xy=2\"$ , and we let $\"z=-1%2F2\"$ , then $\"xyz=2%28-1%2F2%29=-1\"$ which is clearly not less than -1. We must make the requirement that the right side be 'z' and not -1.\r
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\n" ); document.write( "\n" ); document.write( "III. This is true since dividing both sides of an inequality by a negative number will flip the inequality sign. Basically, divide both sides of $\"xy+%3E+1\"$ by the negative number 'z' to get $\"%28xy%29%2Fz+%3C+1%2Fz\"$ (don't forget to flip the sign). \n" ); document.write( "

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