document.write( "Question 303713: If x(x - y) = 0 and x > y, which of the following statements must be true?\r
\n" ); document.write( "\n" ); document.write( "I. x = 0\r
\n" ); document.write( "\n" ); document.write( "II. y < 0\r
\n" ); document.write( "\n" ); document.write( "III. y - x > 0\r
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\n" ); document.write( "\n" ); document.write( "(A) I only\r
\n" ); document.write( "\n" ); document.write( "(B) II only\r
\n" ); document.write( "\n" ); document.write( "(C) I and II only\r
\n" ); document.write( "\n" ); document.write( "(D) I and III only\r
\n" ); document.write( "\n" ); document.write( "(E) I, II, and III \n" ); document.write( "

Algebra.Com's Answer #217621 by jim_thompson5910(35256) $\"\"$ $\"About$

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Well if x > y, then $\"x%3C%3Ey\"$ . So if $\"x%28x+-+y%29+=+0\"$ , then either $\"x=0\"$ or $\"x-y=0\"$ giving $\"x=y\"$ . But we just said that $\"x%3C%3Ey\"$ . So we can only say that $\"x=0\"$ . So statement I is true.\r
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\n" ); document.write( "\n" ); document.write( "If $\"x=0\"$ , then just plug this value into $\"x%3Ey\"$ to get $\"0%3Ey\"$ which shows us that statement II is true.\r
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\n" ); document.write( "\n" ); document.write( "Finally, if $\"x%3Ey\"$ , then subtracting 'x' from both sides gets us $\"0%3Ey-x\"$ which is the opposite of $\"y-x%3E0\"$ . Alternatively, assume that $\"y-x%3E0\"$ is true. Solve for y to get $\"y%3Ex\"$ . This is the opposite of $\"x%3Ey\"$ . So statement III is false.\r
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