document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=573803>Question 573803</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find a number such that twice the number increased by seven is the same as four times the number decreased by three. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #368870 by </B><A HREF=/tutors/aboutme.mpl?userid=mathsmiles><B>mathsmiles(68)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathsmiles><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=mathsmiles><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=368870\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's let X be the number we're looking for.\r<BR>\n" );
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document.write( "The problem says if we take twice the number increased by 7 or ... <BR>\n" );
document.write( "2X + 7<BR>\n" );
document.write( "it should equal four times the number decreased by 3 ... <BR>\n" );
document.write( "4X - 3\r<BR>\n" );
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document.write( "Let's find it!\r<BR>\n" );
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document.write( "2X + 7 = 4X - 3<BR>\n" );
document.write( "Subtract 2X from each side:<BR>\n" );
document.write( "7 = 2X - 3<BR>\n" );
document.write( "Add 3 to each side:<BR>\n" );
document.write( "10 = 2X<BR>\n" );
document.write( "Divide both sides by 2<BR>\n" );
document.write( "5 = X\r<BR>\n" );
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document.write( "Checking:<BR>\n" );
document.write( "2(5) + 7 = 4(5) - 3?<BR>\n" );
document.write( "10 + 7 = 20 - 3?<BR>\n" );
document.write( "17 = 17  Correct! </SPAN>\n" );
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