Question 969531
I just posted this equation, but I don't think it was very clear soo hopefully this format clears up any confusion. I'm taking a course online, and I couldn't be more confused. There's no teacher, and I have no idea where to even start. If someone could show me the steps to solving these equations with the 2 examples I provided, I would really appreciate it. Thanks in advance! 

They ask me to simplify and state the restrictions on the variable. (Dividing/Subtracting Rational Expressions)

1. The numerator consists of x^2 - x-6, and the denominator is x^2 -4. Then we have to DIVIDE that by the numerator which is x^2 -2x+1, divided by the denominator which is x^2-1.


2. The numerator consists of 2x, and is divided by 9x^2+15x. Then we have to SUBTRACT the numerator which is 8, by the denominator which is 3x+5. 
 
I really hope this is clear. Im not sure how else to break it down since I have no idea how to type up equations. Thanks again!


1.
<pre>{{{(x^2 - x - 6)/(x^2 - 4))}}}{{{"÷"}}}{{{(x^2 - 2x + 1)/(x^2 - 1)}}}
{{{(x - 3)(x + 2)cross((x^2 - x - 6))/(x - 2)(x + 2)cross((x^2 - 4))}}}{{{"÷"}}}{{{(x - 1)(x - 1)cross((x^2 - 2x + 1))/(x - 1)(x + 1)cross((x^2 - 1))}}} ------ Factoring all numerators and denominators
{{{(x - 3)(x + 2)/(x - 2)(x + 2)}}}{{{"÷"}}}{{{(x - 1)(x - 1)/(x - 1)(x + 1)}}}
{{{(x - 3)cross((x + 2))/(x - 2)cross((x + 2))}}}{{{"÷"}}}{{{cross((x - 1))(x - 1)/cross((x - 1))(x + 1)}}} ------- Cancelling binomials to make process easier 
{{{(x - 3)/(x - 2)}}}{{{"÷"}}}{{{(x - 1)/(x + 1)}}}
{{{(x - 3)/(x - 2)}}}{{{"*"}}}{{{(x + 1)/(x - 1)}}} ------------ Keep, change, flip [Keep 1st fraction, change the division to multiplication, and then
                           INVERT (flip) fraction after division sign]
{{{((x - 3)(x + 1))/((x - 2)(x - 1))}}}, or {{{highlight_green((x^2 - 2x - 3)/(x^2 - 3x + 2))}}}
2.
{{{2x/(9x^2 + 15x) - 8/(3x + 5)}}} 
{{{2x/3x(3x + 5)cross((9x^2 + 15x)) - 8/(3x + 5)}}} ------ Factoring all numerators and denominators
{{{2x/3x(3x + 5) - 8/(3x + 5)}}} 
{{{2cross(x)/3cross(x)(3x + 5) - 8/(3x + 5)}}} -------- Cancelling x in numerator and x in denominator 
{{{2/3(3x + 5) - 8/(3x + 5)}}}
{{{(2 - 8(3))/3(3x + 5)}}} -------- Multiplying by LCD, 3(3x + 5)
{{{(2 - 24)/3(3x + 5)}}} 
{{{highlight_green((- 22)/3(3x + 5))}}} , or {{{highlight_green((- 22)/(9x + 15))}}}