document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=148145>Question 148145</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello, I am trying to figure Algebra out and it's not clicking to well. If anyone could help I would appreciate it.\r<BR>\n" );
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document.write( "Solve by substitution....<BR>\n" );
document.write( "  8x - 4y = -40<BR>\n" );
document.write( "  6x + 38 = y\r<BR>\n" );
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document.write( "Thank you so very much!!!!!!!!!!!!!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #108551 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=108551\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> substitution means you substitute one thing for another in the equation\r<BR>\n" );
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document.write( "we're given that y=6x+38 __ so we can substitute (6x+38) in place of y in the other equation\r<BR>\n" );
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document.write( "8x-4(6x+38)=-40 __ distributing __ 8x-24x-152=-40 __ adding 152 __ -16x=112 __ dividing by -16 __ x=-7\r<BR>\n" );
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document.write( "substituting to find y __ y=6(-7)+38 __ y=-42+38 __ y=-4 </SPAN>\n" );
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