document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1121723>Question 1121723</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The amount of oil imported to Country A from Country B in millions of barrels per day can be approximated by the equation yequals0.061 x plus 1.23&#8203;, where x is the number of years since 1975. Solve this equation for x. Use the new equation to determine in which year the approximate<BR>\n" );
document.write( "number of oil barrels imported from Country B per day will be 3.06 million. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #737699 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=737699\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=0.061x+1.23<BR>\n" );
document.write( "3.06 is y<BR>\n" );
document.write( "3.06=0.06x+1.23<BR>\n" );
document.write( "1.83=0.06x<BR>\n" );
document.write( "divide by 0.06 both sides<BR>\n" );
document.write( "x=30.5 years which would be 2005.5, or mid-2006, depending upon how 1975 was exactly defined. I define it as ending just before 1976, so 30 years would be just before 2006, and 30.5 in mid-2006. </SPAN>\n" );
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