Question 147022
A: {{{(4/(x-2))+(3/(x^2-x-2)) }}}
Remember, to add or subtract fractions you need a common denominator. 
Luckily, 
{{{x^2-x-2=(x-2)(x+1)}}}
So you can multiply, the first term by (x+1)/(x+1) to get it done.
A: {{{(4/(x-2))+(3/(x^2-x-2)) }}}
A: {{{(4(x+1)/((x-2)*(x+1)))+(3/((x-2)*(x+1)))}}}
A: {{{(4x+4+3)/((x-2)*(x+1)) }}}
A: {{{(4x+7)/((x-2)*(x+1)) }}}
B: {{{4y/(y^2+6y+5)-2y/(y^2-1)}}}
Let's look at both denominators and factor them. 
First term denominator:
{{{y^2+6y+5=(y+5)(y+1)}}}
Second term denominator:
{{{y^2-1=(y+1)(y-1)}}}
Multiply the first term by (y-1)/(y-1) and
then multiply the second term by (y+5)/(y+5) to get
the common denominator (y+1)(y-1)(y+5).
B: {{{4y(y-1)/((y+1)(y+5)(y-1))-2y(y+5)/((y+1)(y-1)(y+5))}}}
B: {{{(4y^2-4y-2y^2-10y)/((y+1)(y-1)(y+5))}}}
B: {{{(2y^2-14y)/((y+1)(y-1)(y+5))}}}
B: {{{2y(y-7)/((y+1)(y-1)(y+5))}}}