document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=387358>Question 387358</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given the linear equation y=-2/3x+1, find the y-coordinates of the points (-6,  ), (-3,  ), and (3,  ). Please show all of your work. Plot those points and graph the linear equation.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #275168 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=275168\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given the linear equation y=-2/3x+1, find the y-coordinates of the points (-6,  ), (-3,  ), and (3,  ). Please show all of your work. Plot those points and graph the linear equation.<BR>\n" );
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document.write( "Assuming you mean y=(-2/3)x+1  and not -2/(3x) + 1 or -2/(3x+1):<BR>\n" );
document.write( "y=(-2/3)x+1<BR>\n" );
document.write( "Sub 6 for x and find y<BR>\n" );
document.write( "y = (-2/3)*6 + 1 = -3 --> (6,-3)<BR>\n" );
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document.write( "x = -3:<BR>\n" );
document.write( "y = 2 + 1 = 3 --> (-3,3)<BR>\n" );
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document.write( "do the same for x = 3<BR>\n" );
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document.write( "Plot the points, draw a line thru them.\r<BR>\n" );
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