```Question 81677
All of this math is new to me, I need some serious help in explaining the steps that lead to the solution. Thanks
<pre><font size = 3><b>
Write down what you're adding something to, namely,

{{{(2y)/(y^2-1)}}}

+ {{{(4y)/(y^2+6y+5)}}} after it, so you have:

{{{2y/(y^2-1)}}} + {{{(4y)/(y^2+6y+5)}}}

Now you factor the denominators:

{{{(2y)/((y-1)(y+1))}}} + {{{(4y)/((y+5)(y+1))}}}

The LCD must contain each factor as many times as
it occurs in any ONE denominator.  Each of those
factors {{{(y-1)}}}, {{{(y+1)}}}, and {{{(y+5)}}}
occurs no more than once in any ONE denominator,
so the LCD will contain them all just ONCE:
LCD = {{{ (y-1)(y+1)(y+5) }}}.

Now look at the first fraction you are to add.

{{{(2y)/((y-1)(y+1))}}}

Ask yourself: "What factor does that denominator need
in order to become the LCD of {{{ (y-1)(y+1)(y+5) }}}?

You look and see that it is lacking the factor {{{(y+5)}}}

So you stick on a {{{(y+5)}}} factor in the top and the
bottom like this:

{{{(2y(y+5))/((y-1)(y+1)(y+5))}}}

Now you do the same with the second fraction you are to add:

{{{(4y)/((y+5)(y+1))}}}

Ask yourself: "What factor does that denominator need
in order to become the LCD of {{{ (y-1)(y+1)(y+5) }}}?

You look and see that it is lacking the factor {{{(y-1)}}}

So you stick on a {{{(y-1)}}} factor in the top and the
bottom like this:

{{{(4y(y-1))/((y+5)(y+1)(y-1))}}}

Now put a + sign between them again:

{{{(2y(y+5))/((y-1)(y+1)(y+5))}}} + {{{(4y(y-1))/((y+5)(y+1)(y-1))}}}

Multiply the tops out but DO NOT multiply the bottoms out.

{{{(2y^2+10y)/((y-1)(y+1)(y+5))}}} + {{{(4y^2-4y)/((y+5)(y+1)(y-1))}}}

Since the denominators are the same LCD (except for the
order of factors which doesn't matter) you can combine the
numerators over the LCD, using the same sign between them
as the sign between the fractions:

{{{((2y^2+10y)+(4y^2-4y))/((y-1)(y+1)(y+5))}}}

Remove the parentheses in the top

{{{(2y^2+10y+4y^2-4y)/( (y-1)(y+1)(y+5) )}}}

Combine like terms in the top:

{{{(6y^2+6y)/( (y-1)(y+1)(y+5) )}}}

Factor out {{{6y}}} in the top:

{{{(6y(y+1))/( (y-1)(y+1)(y+5) )}}}

Cancel the {{{y+1)}}}'s in the top and bottom,
and that leaves just this:

{{{6y/( (y-1)(y+5) )}}}

If you like you can multiply the bottom out,
and leave it as

{{{6y/(y^2+4y-5)}}}

but that is not necessary.

Edwin</pre>```