1. ## Simply help.

The question is simplfy: $\frac{t^2-9}{6-2t} \times \frac{6}{t-3}$

i simply till $\frac{3(t-3)}{3-t)}$ then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is $\frac{3(t-3)}{3-t)}$ . but i can get -3. which shld be the correct ans?

2. $\frac{t^2-9}{6-2t}\cdot\frac{6}{t-3}=\frac{6(t-3)(t+3)}{-2(t-3)^2}=\frac{3(t+3)}{3-t}$

3. Originally Posted by helloying
The qn is simply $\frac{t^2-9}{6-2t} \times \frac{6}{t-3}$

i simply till $\frac{3(t-3)}{3-t)}$ then i mulitpy by $-1$ for both denominator and numerator so i get $-3$ in the end. is this correct? the ans given is $\frac{3(t-3)}{3-t)}$. but i can get $-3$. which shld be the correct ans?
$\frac{t^2-9}{6-2t} \times \frac{6}{t-3}=\frac{(t-3)(t+3)}{2(3-t)}\times \frac{2 \times 3}{t-3}$

Now cancel where possible to get the answer.

CB

4. Hello helloying!

Originally Posted by helloying
The qn is simply $\{frac{\t^2-9}{6-2t} x \frac{\6}{t-3}[\math]

i simply till \frac{\3(t-3)}{3-t)} then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is \frac{3(t-3)}{3-t)}. but i can get -3. which shld be the correct ans?$
Edit: whoops - wasn't the right the solution, Sorry!

Rapha