document.write( "Question 31529: Simplify\r
\n" ); document.write( "\n" ); document.write( "25x2 - 9y2/xy
\n" ); document.write( "-------------- divide by
\n" ); document.write( "5/y - 3/x \n" ); document.write( "

Algebra.Com's Answer #18191 by benni1013(206) $\"\"$ $\"About$

You can put this solution on YOUR website!
First step is to find the common denominators of both the numerator and denominator:
\n" ); document.write( "(25x^2-9y^2)/xy/(5/y-3/x)\r
\n" ); document.write( "\n" ); document.write( "Numerator LCD: xy
\n" ); document.write( "Denominator LCD: xy\r
\n" ); document.write( "\n" ); document.write( "Whatever is missing needs to multiplied:
\n" ); document.write( "The xy is missing below the 25x^2 in the numerator
\n" ); document.write( "The x is missing below the 5 and the y is missing below the 3 so multiply them to the numerator\r
\n" ); document.write( "\n" ); document.write( "Above shown:
\n" ); document.write( "((25x^2(xy)/xy)-9y^2/xy)/(5(x)/xy-3(y)/xy)\r
\n" ); document.write( "\n" ); document.write( "Combine and Simplify
\n" ); document.write( "(25x^3y-9y^2/xy)/(5x-3y/xy)\r
\n" ); document.write( "\n" ); document.write( "Divide:
\n" ); document.write( "(25x^3y-9y^2/xy)(xy/5x-3y)\r
\n" ); document.write( "\n" ); document.write( "The xy's cross cancel to simplify to:
\n" ); document.write( "25x^3y-9y^2/5x-3y\r
\n" ); document.write( "\n" ); document.write( "One last factoring:
\n" ); document.write( "y(25x^3-9y)/5x-3y\r
\n" ); document.write( "\n" ); document.write( "Note: The difference of two square rules doesn't apply here because both of the quantities don't have powers of two. \n" ); document.write( "

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