document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=164775>Question 164775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the sum of two numbers is -3. if the second number is subtracted from the first, the difference is 27. find the numbers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #121424 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=121424\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> so lets call the 2 numbers a and b.  a+b= -3 and a-b= -27 now solve by substitution or similtaneously.  I choose substitution  so solving for the first equation a=-3-b  substituting this into the second equation we get -3-b-b=-27  adding 3 to both sides and combining like b terms gives us -2b=-24 dividing both sides by -2 gives us b=12   no plugging the value of b into either equation lets take the first a+12= -3 after subtracting 12 from both sides we get a= -15<BR>\n" );
document.write( "so a= -15<BR>\n" );
document.write( "   b= 12\r<BR>\n" );
document.write( "\n" );
document.write( "checking by substituting these back into the original equations we see that we have found the correct answers. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );