document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=236513>Question 236513</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If 5 times the first number plus three times the second number equals 47, and 10 times the first number minus 4 times the second number equals 54, what are 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 #174091 by </B><A HREF=/tutors/aboutme.mpl?userid=Anthea+Lawn><B>Anthea Lawn(22)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Anthea+Lawn><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=174091\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> okay, call the first number x and the second number y\r<BR>\n" );
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document.write( "5x + 3y = 47<BR>\n" );
document.write( "10x -4y = 54\r<BR>\n" );
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document.write( "So now we have two simultaneous equations.  Double everything in the top equation :\r<BR>\n" );
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document.write( "10x + 6y = 94\r<BR>\n" );
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document.write( "Now both equations start with a 10x   So take the second equation away from the new top one and the 10x bits will disappear :\r<BR>\n" );
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document.write( "(10x - 10x) + 6y - (-4y) = 94 - 54\r<BR>\n" );
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document.write( "10y = 40   (because taking away a minus equals a plus)\r<BR>\n" );
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document.write( "y = 4\r<BR>\n" );
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document.write( "Substituting back into either of the original equations gives x = 7\r<BR>\n" );
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document.write( "So the two numbers are 7 and 4. </SPAN>\n" );
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