Question 78035
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{{{(2-x/y)/(1/y+y)}}}

Write the {{{2}}} over {{{1}}}} and the {{{y}}} over {{{1}}}
so the whole "building" will be 4 "stories"
high.

{{{(2/1-x/y)/(1/y+y/1)}}}

Put parentheses around the numerator and the
denominator:

{{{((2/1-x/y))/((1/y+y/1))}}}

Look at all four denominators, {{{1}}},{{{y}}},{{{y}}},and {{{1}}}
Their LCD is {{{y}}}. So multiply by {{{y/y}}}
which just equals {{{1}}}. But make it four "stories" 
high also by putting each of the {{{y}}}'s over
{{{1}}} like this

{{{(y/1)/(y/1)}}}

Now since this just equals to {{{1}}}, we can multiply
it by our example without changing its value.

{{{(y/1)/(y/1)}}}{{{((2/1-x/y))/((1/y+y/1))}}} 

Now remove the parentheses by distributing:

{{{( (y/1)(2/1)-(y/1)(x/y) )/( (y/1)(1/y)+(y/1)(y/1) )}}}

The first term on top just becomes {{{2y/1}}} which is
just {{{2y}}}

{{{( 2y-(y/1)(x/y) )/( (y/1)(1/y)+(y/1)(y/1) )}}}

The {{{y}}}'s cancel in the second term on top, leaving
just {{{x/1}}} which is just {{{x}}}

{{{( 2y-x)/( (y/1)(1/y)+(y/1)(y/1) )}}}

The {{{y}}}'s cancel in the first term on the bottom,
leaving {{{1/1}}} which is just {{{1}}}

{{{( 2y-x)/( 1+(y/1)(y/1) )}}}

Finally the second term on the bottom becomes {{{y^2/1}}}
which is just {{{y^2}}} and so we end up with just

{{{( 2y-x)/( 1+y^2 )}}}

Edwin</pre></b></font>