how do i solve: 1/b-a + 1/a+b and 3y-1/2y^2+y-3 - 2-y/y-1 - y/1-yalgebra

You cannot solve these, only simplify.  I see that you use
spaces, but it is easier to undersand if you use
parenthesis.

I am suspecting you
mean

(a) 1/(b-a) + 1/(a+b) -
2b/(a^2-b^2)

(b) (3y-1)/(2y^2+y-3) - (2-y)/(y-1) -
y/(1-y)

(a) First note that 1/(b-a) = -1/(a-b) since (b-a)
= -(a-b)
Now we can factor (a^2-b^2) = (a-b)(a+b) and our expression
is

-1/(a-b) + 1/(a+b) -
2b/((a+b)/(a-b))

Now like all fractions we can only add if
we have a common denominator so I am going to multiply -1/(a-b) * (a+b)/(a+b).  I can do
this because (a+b)/(a+b)=1 and I can always multiply by 1.  With the same idea I can
multiply 1/(a+b) by (a-b)/(a-b) again because (a-b)/(a-b) = 1. so I
get

(-1(a+b))/((a-b)(a+b)) + (1(a-b)/((a-b)(a+b)) -
2b((a+b)(a-b))

Please verify that if you simplify
everything I get the original equation.  Now that I have the same denominator, I add the
numerators, just like (1/4+2/4 = 3/4).

(-1(a+b) + 1(a-b) -
2b)/((a+b)(a-b))  Now use the distributive property to simplify the
numerator.

(-a + -b + a + -b + -2b)/((a+b)(a-b)) =
(-4b)/((a+b)(a-b)) = -4b/((a+b)(a-b)).

So 1/(b-a) + 1/(a+b)
- 2b/(a^2-b^2) = -4b/((a+b)(a-b))

(b) (3y-1)/(2y^2+y-3) -
(2-y)/(y-1) - y/(1-y)  first factor,
2y^2 + y - 3 = (2y+3)(y-1). and (1-y) =
-(y-1) so y/(1-y) = -y/(y-1). this gives
us

(3y-1)/((2y+3)(y-1)) - (2-y)/(y-1) - (-y/(y-1)) Now we
need a common denominator which is (2y+3)(y-1) so multiply the last two rational
expressions with (2y+3)/(2y+3) to get

(3y-1)/((2y+3)(y-1))
- ((2-y)(2y+3))/((y-1)(2y+3)) - (-y(2y+3)/((y-1)(2y+3))

Now
we have the same denominator we can add the numerators
again

((3y-1) - (2-y)(2y+3) - (-y(2y+3)))/((y-1)(2y+3)) 
Now distribute to get

(3y - 1 - (4y - 2y^2 + 6 - 3y) -
(-2y^2 - 3y))/((y-1)(2y+3))  Now change subtractions to additions by changing the sign
of what we are subtracting.

(3y +  -1 + -4y + 2y^2 + -6 +
3y + 2y^2 + 3y)/((y-1)(2y+3)) now add like terms to
get

(4y^2 + 5y -
7)/((y-1)(2y+3))

So our answers
are

(a) 1/(b-a) + 1/(a+b) -
2b(a^2-b^2) = -4b/(a^2-b^2) =
4b/(b^2-a^2)

(b)
(3y-1)/(2y^2+y-3) - (2-y)/(y-1) - y/(1-y) = (4y^2 + 5y -
7)/((y-1)(2y+3))