document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=142850>Question 142850</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 3  1/4y  -  7  1/2  =  2  2/3\r<BR>\n" );
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document.write( "7/x - 2  =  4/x + 3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #103953 by </B><A HREF=/tutors/aboutme.mpl?userid=Fruglemiester1234><B>Fruglemiester1234(13)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fruglemiester1234><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=103953\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I assume you are just trying to find X and Y in these equations.\r<BR>\n" );
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document.write( "1) For this problem I would first of all get the mixed fractions into single fractions(not sure their specific name) Like:<BR>\n" );
document.write( "    13/4y - 15/2 = 8/3        note: you can do this by times the outside # by the denominator then adding the numerator....3 1/4 ... 3*4 +1 =13 ...13/4\r<BR>\n" );
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document.write( "  Then I would get them all to have the same denominator(bottom #)...<BR>\n" );
document.write( "     39/12y - 90/12 = 32/12      note: the least common denominator for 4,3,2 is 12. Times both the 13 and the 4 by 3, you get 39 for the top and 12 for the bottom, do this for all three.\r<BR>\n" );
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document.write( "  Add 90/12 to both sides. You get:<BR>\n" );
document.write( "      39/12y = 32/12 + 90/12<BR>\n" );
document.write( "  You can add 32 and 90, because they both have the same bottom #. you get:<BR>\n" );
document.write( "      39/12y = 122/12<BR>\n" );
document.write( "  Divide both sides by 39/12. Instead of dividing on the other side you can times by the recipricol, as follows:<BR>\n" );
document.write( "      y = 122/12  / 39/12 ... y = 122/12 * 12/39... y = 122*12/39*12<BR>\n" );
document.write( "  12's end up cancelling, because theres one on the top and one on the bottom (like 12/12 = 1). After this you would try to reduce the fraction of 122/39, but in this case it is as simple as it can get.\r<BR>\n" );
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document.write( "y = 122/39 \r<BR>\n" );
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document.write( "or if you want decimal\r<BR>\n" );
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document.write( "y = 3.13\r<BR>\n" );
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document.write( "2)  7/x - 2 = 4/x +3\r<BR>\n" );
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document.write( "  First I would try to get the x's together on one side. We can start to do this by adding 2 to both sides. we end up with:<BR>\n" );
document.write( "       7/x = 4/x + 5<BR>\n" );
document.write( "  Then I would subtract (4/x) from both sides. <BR>\n" );
document.write( "       7/x - 4/x = 5<BR>\n" );
document.write( "  Sense both fractions have the same denominator we can add(in this case subtract) the numerators and write them as one fractions, like this:<BR>\n" );
document.write( "      (7-4)/x = 5 ...7 subtract 4 gets us to..<BR>\n" );
document.write( "       3/x =5<BR>\n" );
document.write( "  Then we times both sides by x, getting the x out from beneath the 3.We get:<BR>\n" );
document.write( "      3 = 5x<BR>\n" );
document.write( "  From here its pretty basic, we divide by 5, getting us:<BR>\n" );
document.write( "      3/5 = x\r<BR>\n" );
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document.write( "X = 3/5\r<BR>\n" );
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document.write( "or in decimal value\r<BR>\n" );
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document.write( "X = .6\r<BR>\n" );
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document.write( " </SPAN>\n" );
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