document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185925>Question 1185925</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the area bounded by the equation x/- 5 + y/2 = 1 on the first quadrant and the line x = 10  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849508 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849508\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to find the area:\r<BR>\n" );
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document.write( "1. **Rewrite the equation:**<BR>\n" );
document.write( "   The equation x/-5 + y/2 = 1 can be rewritten as:<BR>\n" );
document.write( "   y/2 = 1 + x/5<BR>\n" );
document.write( "   y = 2 + (2/5)x\r<BR>\n" );
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document.write( "2. **Find the intersection points:**\r<BR>\n" );
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document.write( "   * **Intersection with the x-axis:**  Set y = 0 in the original equation:<BR>\n" );
document.write( "     x/-5 + 0/2 = 1<BR>\n" );
document.write( "     x/-5 = 1<BR>\n" );
document.write( "     x = -5<BR>\n" );
document.write( "     This point is (-5, 0). However, we are only interested in the first quadrant, where x and y are both positive.\r<BR>\n" );
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document.write( "   * **Intersection with the y-axis:** Set x = 0 in the original equation:<BR>\n" );
document.write( "     0/-5 + y/2 = 1<BR>\n" );
document.write( "     y/2 = 1<BR>\n" );
document.write( "     y = 2<BR>\n" );
document.write( "     This point is (0, 2).\r<BR>\n" );
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document.write( "   * **Intersection with the line x = 10:** Substitute x = 10 into the equation:<BR>\n" );
document.write( "     10/-5 + y/2 = 1<BR>\n" );
document.write( "     -2 + y/2 = 1<BR>\n" );
document.write( "     y/2 = 3<BR>\n" );
document.write( "     y = 6<BR>\n" );
document.write( "     This point is (10, 6).\r<BR>\n" );
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document.write( "3. **Visualize the area:** The area we're looking for is a trapezoid. The vertices of the trapezoid are (0, 2), (10, 6), and the points where x=10 intersects the x-axis (10,0).\r<BR>\n" );
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document.write( "4. **Calculate the area of the trapezoid:** The formula for the area of a trapezoid is:\r<BR>\n" );
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document.write( "   Area = (1/2) * (base1 + base2) * height\r<BR>\n" );
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document.write( "   In our case:<BR>\n" );
document.write( "   * base1 = the length of the vertical line at x=0, which is 2<BR>\n" );
document.write( "   * base2 = the length of the vertical line at x=10, which is 6<BR>\n" );
document.write( "   * height = the horizontal distance between the two vertical lines, which is 10\r<BR>\n" );
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document.write( "   Area = (1/2) * (2 + 6) * 10<BR>\n" );
document.write( "   Area = (1/2) * 8 * 10<BR>\n" );
document.write( "   Area = 40\r<BR>\n" );
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document.write( "Therefore, the area bounded by the given equation, the x and y axes in the first quadrant, and the line x = 10 is $\boxed{40}$ square units.<BR>\n" );
document.write( " </SPAN>\n" );
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