hello, i cannot do 10 b and c, the image above shows how much ive gotten done, for b) im not sure how to even start or how i can work through this problem; never seen it before
Lets look at b)
2x/(x^3 +x^2 - 6x) - (x-8)/(x^2 - 5x - 24)
= 2x/[x*(x^2 + x - 6)] - (x-8)/[(x-8)(x+3)]
= 2/[x^2 + x - 6] - 1/(x+3)
= 2/[(x+3)(x-2)] - 1/(x+3)
= 2/[(x+3)(x-2)] - (x-2)/[(x-2)(x+3)]
= (4 - x)/[(x+3)(x-2)] (since 2 - x + 2 = 4 - x).
Deal with the "biggest" fraction by working on its numerator, then working on its denominator.
How could you simplify $\displaystyle \displaystyle \begin{align*} \frac{3}{2} + \frac{3}{t} \end{align*}$? How would you simplify $\displaystyle \displaystyle \begin{align*} \frac{t}{t + 6} - \frac{1}{t} \end{align*}$?
It's not $\displaystyle \displaystyle \begin{align*} \frac{3t(6)}{2t} \end{align*}$, I can see you're trying to apply a rule $\displaystyle \displaystyle \begin{align*} \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \end{align*}$ without really understanding what's going on (and messing up the numerator).
To add two fractions, you need a common denominator. In the first your denominator is 2, and in the second your denominator is t. So the common denominator is 2t. Whatever you do to the bottom of each fraction to get the common denominator, you need to do to the top as well. You're really multiplying each fraction by a cleverly disguised 1, because when you multiply by 1, your number remains the same. So
$\displaystyle \displaystyle \begin{align*} \frac{3}{2} + \frac{3}{t} &= \frac{3}{2} \cdot \frac{t}{t} + \frac{3}{t} \cdot \frac{2}{2} \textrm{ (do you see how each of these fractions I've multiplied by is really 1?)} \\ &= \frac{3t}{2t} + \frac{6}{2t} \\ &= \frac{3t + 6}{2t} \end{align*}$