Question 203264
The key is to multiply the numerator and denominator of the "big" fraction by the Lowest Common Denoinator (LCD) of the "small" fractions in that "big" fraction.<br>
{{{(a/b+c)/(a/b-c)}}}
The LCD of the "small" fractions is b
{{{((a/b+c)/(a/b-c))*(b/b) = (a + bc)/(a - bc)}}}<br>
{{{(1/4+x)/(1/2)}}}
The LCD is 4
{{{((1/4+x)/(1/2))*(4/4) = (1 + 4x)/2 = 1/2 + 2x}}}
Since the denominator is just a single term we might also change the "divide by 1/2" to "multiply by the reciprocal of 1/2". The reciprocal of 1/2 is 2/1 = 2.:
{{{(1/4+x)/(1/2) = (1/4+x)*2 = 1/2 + 2x}}}<br>
{{{(1/3+1/5-1/7)/(1/x)}}}
The LCD is 105x
{{{((1/3+1/5-1/7)/(1/x))*((105x)/(105x)) = (35x + 21x - 15x)/105 = 41x/105}}}
Since the denominator is just a single term we might also change the "divide by 1/x" to "multiply by the reciprocal of 1/x". The reciprocal of 1/x is x/1 = x.:
{{{((1/3+1/5-1/7)/(1/x)) = (1/3+1/5-1/7)*x = x/3 + x/5 - x/7}}}. If you then get the denominators the same and add and subtract you get the answer above: 41x/105<br>
{{{(a/b+2+b/a)/(a/b-b/a)}}}
The LCD is ab
{{{((a/b+2+b/a)/(a/b-b/a))*((ab)/(ab)) = (a^2 + 2ab + b^2)/(a^2 - b^2)}}}
While none of the previous answers would reduce, this one will. We reduce by factoring and canceling common factors:
{{{(a^2 + 2ab + b^2)/(a^2 - b^2) = ((a + b)(a + b))/((a + b)(a - b))}}}
One of the (a+b)'s in the numerator cancels with the one in the denominator leaving:
{{{(a + b)/(a - b)}}}