document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=5698>Question 5698</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I actually have 2 questions relating to adding and subtracting fractions using variables.<BR>\n" );
document.write( "1)x-y/x+y  +  x+y/x-y\r<BR>\n" );
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document.write( "I thought it was 1 but now it seems that would be too easy.\r<BR>\n" );
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document.write( "2) 1/x - 1/x+4 = 1/3\r<BR>\n" );
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document.write( "*I know I'm supposed to find the LCD and multiply both sides by it, but I'm not sure what that is.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2951 by </B><A HREF=/tutors/aboutme.mpl?userid=guapa><B>guapa(62)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=guapa><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=2951\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1) You have to find the LCD which is (x+y)(x-y). Now you have to multiply each fraction with the missing term:<BR>\n" );
document.write( "(x-y)(x-y)/(x+y)(x-y) + (x+y)(x+y)/(x-y)(x+y)<BR>\n" );
document.write( "With a common denominator in both fractions we can rewrite as follows:<BR>\n" );
document.write( "(x-y)(x-y)+(x+y)(x+y)/(x+y)(x-y) Now let's simplify the numerator<BR>\n" );
document.write( "(x^2-2xy+y^2)+(x^2+2xy+y^2)/(x+y)(x-y) Combine like terms in the numerator<BR>\n" );
document.write( "(2x^2+2y^2)/(x+y)(x-y)<BR>\n" );
document.write( "This is already your final answer. You can factor out the 2 in the numerator and evaluate the denominator but it won't change the value.<BR>\n" );
document.write( "2(x^2+y^2)/(x^2-y^2) The sum of two squares is not factorable unless there is a common factor. \r<BR>\n" );
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document.write( "2)Like you already said you have to find the LCD which is 3x(x+4) in this case. Let's look how we found the LCD. Look at every denominator: x , (x+4) , 3<BR>\n" );
document.write( "The LCD contains each different (prime) factor the most times it appears. Thus 3x(x+4).Now we have to multiply each fraction(numerator and denominator) by the missing term to produces equivalent fractions with the same denominator.<BR>\n" );
document.write( "1/x-1/(x+4) = 1/3<BR>\n" );
document.write( "{1*3(x+4)}/{x*3(x+4) - {1*(3x)}/{(x+4)*3x} = {1*x(x+4)}/{3*x(x+4)}<BR>\n" );
document.write( "A big advantage when computing equations with fractions is that by using the LCD you can clear the equation of all fractions.<BR>\n" );
document.write( "3(x+4) - 3x = x(x+4) Simplify both sides<BR>\n" );
document.write( "3x+12-3x = x²+4x Combine like terms<BR>\n" );
document.write( "x²+4x-12=0 To solve this equation we have to factor out the GCF. First we need to find 2 numbers whose sum is 4 and whose product is -12. 6 and -2 satisfy options. <BR>\n" );
document.write( "x²+6x-2x-12=0 Now we can factor out<BR>\n" );
document.write( "x(x+6)-2(x+6)=0<BR>\n" );
document.write( "(x-2)(x+6)=0<BR>\n" );
document.write( "Solve for x<BR>\n" );
document.write( "x-2=0, x=2<BR>\n" );
document.write( "x+6=0, x=-6<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "1/2 - 1/(2+4) = 1/3<BR>\n" );
document.write( "1/2 - 1/6 = 1/3<BR>\n" );
document.write( "3-1 = 2, 2=2<BR>\n" );
document.write( "and <BR>\n" );
document.write( "1/(-6) - 1/(-6+4) = 1/3<BR>\n" );
document.write( "1/(-6) - 1/(-2) = 1/3<BR>\n" );
document.write( "-1+3=2, 2=2<BR>\n" );
document.write( "Hope it helps<BR>\n" );
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