document.write( "Question 54286: 2 divided by 2/X + 2/Y is problem one...i have an answer guide so i know the answer i just don't know how to do it on my own\r
\n" ); document.write( "\n" ); document.write( "2x^2/3y * y^3/8x is problem two... this too i have the answer to but i don't know how to do it myself
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Algebra.Com's Answer #36566 by jainenderkapoor(61)\"\" \"About 
You can put this solution on YOUR website!
Q.1 2 divided by (2/x + 2/y)
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\n" ); document.write( " (2/x + 2/y) = 2 (1/x + 1/y)\r
\n" ); document.write( "\n" ); document.write( " = 2 ( (y + x)/xy)
\n" ); document.write( " (I solved 1/x + 1/y by finding common denominator xy and then multiplying the numerator by x and y respectively) \r
\n" ); document.write( "\n" ); document.write( " = 2(y+x)/xy\r
\n" ); document.write( "\n" ); document.write( "Now we have to solve 2 divided by 2(y + x)/xy\r
\n" ); document.write( "\n" ); document.write( " we get 2 xy/2(y + x)
\n" ); document.write( " = xy/(y + x)\r
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\n" ); document.write( "\n" ); document.write( "Q.2 (2x^2/3y)* (y^3/8x)
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\n" ); document.write( " Here we will simply multiply the numerators and the denominators\r
\n" ); document.write( "\n" ); document.write( " (2x^2 * y^3)/(3y*8x)\r
\n" ); document.write( "\n" ); document.write( " we get the power of x as 2 - 1 = 1 in the numerator and the power
\n" ); document.write( "of y as 3 - 1 = 2 in the numerator
\n" ); document.write( "We get 2/24 which can be written as 1/12 \r
\n" ); document.write( "\n" ); document.write( "so the final answer becomes \r
\n" ); document.write( "\n" ); document.write( " (x y^2)/12\r
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\n" ); document.write( "\n" ); document.write( "I hope that the solution is clear to you.\r
\n" ); document.write( "\n" ); document.write( "If you have any doubt you are welcome to contact me.\r
\n" ); document.write( "\n" ); document.write( "I am also available for online tutoring in math at very reasonable rates.\r
\n" ); document.write( "\n" ); document.write( "My mail id is jainenderkapoor@gmail.com\r
\n" ); document.write( "\n" ); document.write( "Jainender Kapoor
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