Question 369132
{{{(1/(a+b) - 1/(a-b))/(1/(a-b) + 1/(a+b))}}}
Because subtraction of fractions can cause many errors, I like to change them into additions of the opposite:
{{{(1/(a+b) + (-1)/(a-b))/(1/(a-b) + 1/(a+b))}}}
There are several ways to simplify this expression. The way I like to do this is:<ol><li>Find the Lowest Common Denomintor (LCD) of all the "little" denominators.</li><li>Multiply the numerator and denominator of the "big" fraction by the LCD found in step 1.</li></ol>
Let's see how this works on your problem.
1) Find the LCD of the "little" denominators. The "little" denominators are (a+b) and (a-b). The LCD of these two is simply their product: (a+b)(a-b)
2) Multiply the numerator and denominator of the "big" fraction by the LCD:
{{{((1/(a+b) + (-1)/(a-b))/(1/(a-b) + 1/(a+b)))*(((a+b)(a-b))/((a+b)(a-b)))}}}
In both the numerator and denominator we will need to use the Distributive Property to multiply:
{{{((1/(a+b))*(a+b)(a-b) + ((-1)/(a-b))*(a+b)(a-b))/((1/(a-b))*(a+b)(a-b) + (1/(a+b))*(a+b)(a-b))}}}
Now <i>all</i> the "little" denominators cancel out:
{{{((1/cross((a+b)))*cross((a+b))(a-b) + ((-1)/cross((a-b)))*(a+b)cross((a-b)))/((1/cross((a-b)))*(a+b)cross((a-b)) + (1/cross((a+b)))*cross((a+b))(a-b))}}}
leaving:
{{{(1*(a-b) + (-1)*(a+b))/(1*(a+b) + 1*(a-b))}}}
which simplifies as follows:
{{{(a - b + (-a) + (-b))/(a + b + a - b)}}}
{{{(-2b)/(2a)}}}
The 2's cancel leaving
{{{(-b)/a}}}