document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=693436>Question 693436</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do i do Given the function, (1/3X) + 3/4 = Y, what is Y when X = 1? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #427390 by </B><A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><B>RedemptiveMath(80)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RedemptiveMath><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=427390\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> All we need to do is put 1 wherever we find x in the equation. Once we \"plug\" 1 in for x, we solve using order of operations. So,\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "(1/3X) + 3/4 = Y (writing your equation)<BR>\n" );
document.write( "(1/3)(1) + 3/4 = Y (I'm assuming that 3 in 1/3 and x are being multiplied?)<BR>\n" );
document.write( "1/3 + 3/4 = Y (multiplication before addition)<BR>\n" );
document.write( "4/12 + 9/12 = Y (find common denominators before adding fractions)<BR>\n" );
document.write( "13/12 = Y (add)<BR>\n" );
document.write( "1 1/12 = Y (rewrite improper fraction as mixed number).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "It depends on your teacher whether you keep the fraction improper or write it as a mixed number. You can now say that when x is 1, y is 1 and 1/12 or 13/12.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );