document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=101849>Question 101849</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what if i forgot the value of 8 times 7 how cold i use the distributive propertyto find 8 times seven </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #74043 by </B><A HREF=/tutors/aboutme.mpl?userid=Fombitz><B>Fombitz(32388)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fombitz><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/cgi-bin/show-face.mpl?user=Fombitz><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=74043\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's review the distributive property. <BR>\n" );
document.write( "A*(B+C) = A*B + A*C<BR>\n" );
document.write( "So how does that apply to 8*7?<BR>\n" );
document.write( "8*(7)=8*(6+1)=8*(5+2)=8*(4+3)<BR>\n" );
document.write( "So all of these sums should be the same. Let's check. We'll work backwards.<BR>\n" );
document.write( "8*(4+3) = 8*4 + 8*3 = 32 + 24 = 56<BR>\n" );
document.write( "8*(5+2) = 8*5 + 8*2 = 40 + 16 = 56<BR>\n" );
document.write( "8*(6+1) = 8*6 + 8*1 = 48 +  8 = 56<BR>\n" );
document.write( "So then 8*7 = 56<BR>\n" );
document.write( "Eventually you could get down to <BR>\n" );
document.write( "8*7 = 8*1 + 8*1 + 8*1 + 8*1 + 8*1 + 8*1 + 8*1 + 8*1 or 8 added 7 times or (8 times 7) and that would still equal 56. <BR>\n" );
document.write( "Multiplication is just quicker addition.  </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );