document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1008600>Question 1008600</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A petroleum company has two different sources of crude oil. the first source provides crude oil that is 25% hydrocarbons and the second one provides crude oil that is 50% hydrocarbons. In order to obtain 60 gallons of crude oil that is 30% hydrocarbons how many gallons of crude oil must be used from each of the two sources\r<BR>\n" );
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document.write( "first source = \r<BR>\n" );
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document.write( "Second source = </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #624201 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=624201\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's call the first source x and the second source y:<BR>\n" );
document.write( "x+y= 60 subtract x from both sides:<BR>\n" );
document.write( "y= 60-x<BR>\n" );
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document.write( ".25x+.50y= .30(60)<BR>\n" );
document.write( ".25x+.50(60-x)= .30(60)<BR>\n" );
document.write( ".25x+30-.50x= 18 Now subtract x on left and subtract 30 on both sides:<BR>\n" );
document.write( "-.25x= -12 Divide both sides by -.25, and remember that -/- = +<BR>\n" );
document.write( "x= 48 You need 48 gallons of .25 and 60-48= 12 gallon of .5 </SPAN>\n" );
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