document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=138891>Question 138891</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three batteries are connected in series so that the total voltage is 54 volts. The voltage of the first battery is twice the voltage of the second and 1/3 the voltage of the third battery. Find the actual voltage of each battery. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #101336 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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=101336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> LET X BE THE VALUE OF THE FIRST BATTERY.<BR>\n" );
document.write( "THEN:<BR>\n" );
document.write( "X+X/2+3X=54<BR>\n" );
document.write( "(2X+X+6X)/2=54<BR>\n" );
document.write( "9X=108<BR>\n" );
document.write( "X=108/9<BR>\n" );
document.write( "X=12 VOLTS FOR THE FIRST BATTERY.<BR>\n" );
document.write( "12/2=6 VOLTS FOR THE SECOND BATTERY.<BR>\n" );
document.write( "3*12=36 VOLTE FOT THE THIRD BATTERY.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( "12+6+36=54<BR>\n" );
document.write( "54=54 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );