Question

Help me with this problem!!!

Answer 1

\[\frac{ m-3 }{ 3m-9 }\]

Answer 2

Actually just factorise 3 on the bottom.\[3m-9 = 3(m-3)\]does that make sense?

Answer 3

no, but i dont understand factoring....
Ill trust you though. What next?

Answer 4

But can you see how \[3(m-3)\]when you multiply out becomes\[(3\times m) -(3 \times 3)\]?

Answer 5

yeah...

Answer 6

\[{m-3 \over 3(m-3)}\]We can cancel the (m-3)\[{1\cancel{m-3} \over 3\cancel{(m-3)}}\]

Answer 7

Which leaves\[\frac{1}{3}\]

Answer 8

Hope that helps! Let me know if there is any part you don't get.

Answer 9

Where did you get the one from?

Answer 10

Well we have \[(m-3)\]right? How many (m-3) do we have? We have 1. On the bottom of our equation we have \[3(m-3)\]How many (m-3) do we have? We have 3.

Answer 11

The 1 was always there but there was no need to write it.

Answer 12

In the same way that \[\frac{5}{5} = 1\]

Answer 13

O. Thats one of the things i never got...

Answer 14

Thank you so much!!

Answer 15

No problem!