Question

Please also provide details steps to this answer. PLEASE HELP ME!

Answer 1

Answer:

C

Step-by-step explanation:

First, to make things simpler, let's create a common denominator for every term. We can do this by multiplying (x+4) and (x+3) when the denominator has a (x+3) or (x+4), respectively. In other words:

$(\frac{3}{x+3} -\frac{4}{x+4})\div (\frac{2x}{x+3}-\frac{x}{x+4})$

$(\frac{3(x+4)}{(x+3)(x+4)}-\frac{4(x+3)}{(x+4)(x+3)}) \div (\frac{2x(x+4)}{(x+3)(x+4)} -\frac{x(x+3)}{(x+4)(x+3)} )$

Now that they all have a common denominator, we can combine them:

$(\frac{3(x+4)-4(x+3)}{(x+3)(x+4)})\div (\frac{2x(x+4)-x(x+3)}{(x+3)(x+4)} )$

Now, divide them. Recall how to divide fractions. You "flip" the second term and change the division sign into a multiplication sign.

$=(\frac{3(x+4)-4(x+3)}{(x+3)(x+4)})\cdot (\frac{(x+3)(x+4)}{2x(x+4)-x(x+3)} )$

Notice the denominator of the first term and the numerator of the second; we can cancel them out.

$=\frac{3(x+4)-4(x+3)}{2x(x+4)-x(x+3)}$

Now, we just need to simplify.

$=\frac{(3x+12)+(-4x-12)}{(2x^2+8x)+(-x^2-3x)} =\frac{-x}{x^2+5x} =\frac{-1}{x+5}=-\frac{1}{x+5}$

Answer 2

$\left(\dfrac3{x+3}-\dfrac4{x+4}\right)\div\left(\dfrac{2x}{x+3}-\dfrac x{x+4}\right)=\\\\\\=\left(\dfrac{3(x+4)}{(x+3)(x+4)}-\dfrac{4(x+3)}{(x+4)(x+3)}\right)\div\left(\dfrac{2x(x+4)}{(x+3)(x+4)}-\dfrac{x(x+3)}{(x+4)(x+3)}\right)=\\\\\\=\dfrac{3(x+4)-4(x+3)}{(x+3)(x+4)}\div\dfrac{2x(x+4)-x(x+3)}{(x+3)(x+4)}=\\\\\\=\dfrac{3x+12-4x-12}{(x+3)(x+4)}\times\dfrac{(x+3)(x+4)}{x[2(x+4)-(x+3)]}=\\\\\\=\dfrac{-x}{(x+3)(x+4)}\times\dfrac{(x+3)(x+4)}{x(2x+8-x-3)}=\\\\\\=\dfrac{-1}1\times\dfrac1{x+5}\ =\ -\dfrac1{x+5}$