Question 571082
How do I solve fractions like this: m+n over m-n minus m over m+n minus 2m squared over n squared - m squared ?
:
You can't solve this, you have no equal sign, you can only simplify it.
:
Assuming the problem is:
{{{(m+n)/(m-n)}}} - {{{m/(m+n)}}} - {{{(2m^2)/(n^2 - m^2)}}}
we want to change the signs in the third denominator so we can factor the difference of squares to:
{{{(m+n)/(m-n)}}} - {{{m/(m+n)}}} - {{{(2m^2)/(-1(m^2-n^2))}}}
that changes the sign to:
{{{(m+n)/(m-n)}}} - {{{m/(m+n)}}} + {{{(2m^2)/((m^2-n^2))}}}
Factor
{{{(m+n)/(m-n)}}} - {{{m/(m+n)}}} + {{{(2m^2)/((m+n)(m-n))}}}
you can see we can put this over a single denominator
{{{((m+n)(m+n) - m(m-n) + 2m^2)/((m+m)(m-n))}}} = {{{(m^2+2mn+n^2 - m^2+mn + 2m^2)/((m+m)(m-n))}}}
combine like terms, then factor 
{{{(2m^2 + 3mn + n^2)/((m+m)(m-n))}}} = {{{((2m+n)(m+n))/((m+m)(m-n))}}}
Cancel the (m+n), leaving us with
{{{(2m+n)/(m-n)}}}