Question

Number 2: http://imageshack.us/photo/my-images/7/scannu.png/

Answer 1

@satellite73

Answer 2

number 2, 3, and 4 are similar
you have to find the least common multiple of the denominators

Answer 3

ok..

Answer 4

\[x^2-3x=x(x-3)\] and \[3x-9=3(x-9)\]

Answer 5

no

Answer 6

oh ok

Answer 7

you have to factor each of them

Answer 8

then you have to take each factor that you see and multiply them together

Answer 9

so LCM is \(3x(x-3)\)

Answer 10

\[\frac{4}{x(x-3)}=\frac{4\times 3}{3x(x-3)}\] and
\[\frac{6}{3(x-3)}=\frac{6x}{3x(x-3)}\]

Answer 11

now that the denominators are the same, you can add the numerators
just like with numbers

Answer 12

you get
\[\frac{6x+12}{3x(x-3)}=\frac{2x+4}{x(x-3)}\]

Answer 13

the restriction is x can not equal 3?

Answer 14

typo there sorry
i mean
\[\frac{6}{3(x-3)}=\frac{2}{x-3}\]

Answer 15

yes x cannot be 3, and also x cannot be 0

Answer 16

Can you help me with number 3?

Answer 17

\[2\left( x-1 \right)\]
\[x \left( x-1 \right)\]

Answer 18

is that what you do first for the 3rd one

Answer 19

but wait number 3 isn't adding or subtract