Question 35538
Bare with me and I will try to explain;
{{{(1/x + 1/y)/(1/y -1/x)}}}
When you add or subtract fractions you need LCD;
so we will start with the numerator;
{{{1/x+1/y}}}
multiply the first side by y; and the second by x;
{{{(y)1/(y)x+(x)1/(x)y}}};
{{{y/xy+x/xy}}}
now you have common denominators so you can add;
{{{(y+x)/xy}}}; now do the same thing with the denominator;
1/y -1/x
multiply the first side by x and the second by y;
{{{(x)1/(x)y}}}-{{{(y)1/(y)x}}};
{{{x/xy-y/xy}}}
subtract;
{{{(x-y)/xy}}}
Now lets but the equation back together;
{{{((y+x)/(xy))/((x-y)/xy)}}}
Now, when you divide fractions the second term is inversed(reciprocal), and you mulitply;
{{{((y+x)/xy)}}}*{{{(xy/(x-y))}}}
Now cross cancel;
{{{((x+y)/xy) *(xy/(x-y))}}}
{{{(x+y)/(x-y)}}}
Hope you were able to bare with me and understand
=)