document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=107089>Question 107089</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Someone please help me solve this problem, I am going to have similar ones on a test. Find three consecutive even intergers such that three times the second minus the third is 12 more than the first. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #77911 by </B><A HREF=/tutors/aboutme.mpl?userid=Annabelle1><B>Annabelle1(69)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Annabelle1><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=77911\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let you integers be x, x+1, x+2<BR>\n" );
document.write( "three times the second is 3(x+1) minus the third -(x+2)=12 more than the first (x+1)+12\r<BR>\n" );
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document.write( "so 3(x+1)-(x+2)=(x+1)+12<BR>\n" );
document.write( "3x+3-x-2=x+1+12<BR>\n" );
document.write( "2x+3=x+13\r<BR>\n" );
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document.write( "minus x from both sides<BR>\n" );
document.write( "2x-x+3=x-x+13<BR>\n" );
document.write( "x+3=13\r<BR>\n" );
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document.write( "minus 3 from both sides<BR>\n" );
document.write( "x+3-3=13-3<BR>\n" );
document.write( "x=10\r<BR>\n" );
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document.write( "the first integer is 10 so teh second and third (x+1) and (x+2) are 11 and 12 </SPAN>\n" );
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