document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=625305>Question 625305</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the inverse and check.<BR>\n" );
document.write( "h(x)= 1/3 x + 1\r<BR>\n" );
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document.write( "h(x)= 7/8x-2\r<BR>\n" );
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document.write( "d(x)= 5x+7/2x+4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #393412 by </B><A HREF=/tutors/aboutme.mpl?userid=jankiz><B>jankiz(13)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jankiz><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=jankiz><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=393412\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Always remember, the inverse is a way to undo the equation.  <BR>\n" );
document.write( "First problem: H(x)=1/3x+1.  Which is y=1/3x+1.  You subtract 1 from both sides, then you multiply both sides by three.  So the inverse is: 3(y-1).<BR>\n" );
document.write( "H(x)=7/8x-2.  Which is y=7/8x-2.  First yo add 2 to both sides, then multiply by the reciprocal of 8/7.  You get 8/7(y+2). <BR>\n" );
document.write( "D(x)=5x+7/2x+4.  Combine like terms.  So, y=8.5x+4.  Subtract 4 from both sides, then divide by 8.5. So, (y-4)/(8.5).  </SPAN>\n" );
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