document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=378064>Question 378064</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a cashier has 20 bills, all of which are $10 or $20 bills. The total value of the money is $330. How many of each type does the cashier have? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #268597 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=268597\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let a,b be the number of $10 and $20 bills, respectively. We have the system of equations\r<BR>\n" );
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document.write( "10a + 20b = 330<BR>\n" );
document.write( "a + b = 20 (since there are 20 bills)\r<BR>\n" );
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document.write( "We can multiply the second equation by 10 to obtain 10a + 10b = 200. Substituting into the first equation we have\r<BR>\n" );
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document.write( "(10a + 10b) + 10b = 330<BR>\n" );
document.write( "200 + 10b = 330<BR>\n" );
document.write( "b = 13, a = 7\r<BR>\n" );
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document.write( "So the cashier has seven $10 bills, 13 $20 bills. </SPAN>\n" );
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