document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=209846>Question 209846</A>: <I><SPAN STYLE=\"word-break: break-all;\"> three less than the product of a number and four is -67. find the number </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #161175 by </B><A HREF=/tutors/aboutme.mpl?userid=tsuffern><B>tsuffern(3)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tsuffern><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=161175\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> three less than the product of a number and four is -67. find the number\r<BR>\n" );
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document.write( "Let x=unknown variable <BR>\n" );
document.write( "The product of a number and four would be 4x (product meaning multiplication) you now need 3 less than 4x, which would be 4x-3.  Is -67 means it equals -67.  Equation is:\r<BR>\n" );
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document.write( "4x-3=-67\r<BR>\n" );
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document.write( "Now work the equation to isolate x on one side of the =. Start with adding 3 to both sides to get this 4x-3+3=-67+3.  -3+3=0 so we now have 4x on the left.     -67+3=-64.  we now have -64 on the right.\r<BR>\n" );
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document.write( "4x=-64<BR>\n" );
document.write( "Now divide both sides by 4. 4x/4=x, -64/4=-16 0r X=-16, now go back to original equation and substitute x with -16<BR>\n" );
document.write( "4(-16)-3=-67<BR>\n" );
document.write( "4(-16)=-64<BR>\n" );
document.write( "-64-3=-67<BR>\n" );
document.write( "True<BR>\n" );
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