Question 1129164
{{{(1/x + 7/(x+1))/(7/(x-1) - 1/x)}}}
<pre>
The least common denominator of all three fractions, top and bottom, is {{{x*(x+1)*(x-1)}}}, so
multiply every term in numerator and denominator by {{{((x*(x+1)*(x-1))/1)}}}:

{{{(((x*(x+1)*(x-1))/1)expr(1/x) + ((x*(x+1)*(x-1))/1)expr(7/(x+1)))/(((x*(x+1)*(x-1))/1)expr(7/(x-1)) - ((x*(x+1)*(x-1))/1)expr(1/x))}}}

{{{(((cross(x)*(x+1)*(x-1))/1)expr(1/cross(x)) + ((x*(cross(x+1))*(x-1))/1)expr(7/(cross(x+1))))/(((x*(x+1)*(cross(x-1)))/1)expr(7/(cross(x-1))) - ((cross(x)*(x+1)*(x-1))/1)expr(1/cross(x)))}}}

{{{( (x+1)*(x-1) + 7*x*(x-1)  )/( 7*x*(x+1)- (x+1)*(x-1) ))}}} 

{{{( (x^2-x+x-1) + (7x^2-7x)  )/( (7x^2+7x)- (x^2-x+x-1) ))}}} 

{{{( (x^2-1) + (7x^2-7x)  )/( (7x^2+7x)- (x^2-1) ))}}}

{{{( x^2-1 + 7x^2-7x  )/( 7x^2+7x- x^2+1 ))}}}

{{{(8x^2 - 7x - 1)/(6x^2 + 7x + 1)}}}

We can factor the numerator and the denominator:

{{{((8x + 1) (x - 1))/((6x + 1)(x + 1))}}}

But nothing cancels.

Edwin</pre>