document.write( "Question 534044: The product of second and third of 3 consecutive integers is 3 more than 3 times the second integer. find the 3 integers \n" ); document.write( "
Algebra.Com's Answer #351257 by boilpoil(127)\"\" \"About 
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Let x be the first integer
\n" ); document.write( "Then x+1 be the second
\n" ); document.write( "And x+2 be the third\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28x%2B1%29%28x%2B2%29=3%28x%2B1%29%2B3\"
\n" ); document.write( "\"x%5E2%2B2x%2Bx%2B2=3x%2B3%2B3\"
\n" ); document.write( "\"x%5E2%2B3x%2B2=3x%2B6\"
\n" ); document.write( "\"x%5E2%2B3x-3x=6-2\"
\n" ); document.write( "\"x%5E2=4\"
\n" ); document.write( "\"sqrt%28x%5E2%29=sqrt%284%29\"
\n" ); document.write( "x=+-2 ---Sorry can't enter the +_ Sign\r
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\n" ); document.write( "\n" ); document.write( "So the first integer is either 2 or -2\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The second integer
\n" ); document.write( "=+-2+1
\n" ); document.write( "=3/-1\r
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\n" ); document.write( "\n" ); document.write( "The third integer
\n" ); document.write( "=+-3+1
\n" ); document.write( "=4/0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Filtering out:
\n" ); document.write( "3(4)=3(3)+3
\n" ); document.write( "12=9+3
\n" ); document.write( "12=12
\n" ); document.write( "TRUE\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "3(-1)=3(0)+3
\n" ); document.write( "-3=0+3
\n" ); document.write( "-3=3
\n" ); document.write( "FALSE\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So the three integers are 3, 4 and 5
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