Question

another

Answer 1

\[\frac{ 5x }{ x ^{2}-9 }+\frac{ 7 }{ x+3 }\]

Answer 2

Make common denominator of x^2 -9

Answer 3

ok so it shoud be \[\left( x+3 \right)\left( x-3 \right)\]

Answer 4

yeah, thats x^2-9
now what will you do to make the denominator of 7 also
(x+3)(x-3) ?

Answer 5

well i got for the whole thing \[\frac{ 2x ^{2}+6 }{(x+3)(x-3) ? }\]

Answer 6

lets check :
\(\large \frac{ 5x }{ x ^{2}-9 }+\frac{ 7 }{ x+3 }\times \frac{x-3}{x-3}=\frac{5x+7(x-3)}{(x+3)(x-3)}\)
i don't think that will simplify to what u have written...

Answer 7

or is it 12x

Answer 8

whats 5x+7(x-3) = ?

Answer 9

ya it was 12xi got not 2x

Answer 10

\[5x ^{2}-15+7x-21\]

Answer 11

why do u have 5x^2-15 ??
u don't need to multiply 5x with x-3

Answer 12

as the denom. already is (x+3)(x-3)
so the numerator stays as 5x only

Answer 13

oh ok so 5x+7x-21

Answer 14

yeah, which is?

Answer 15

\[12x ^{2}-21\]

Answer 16

5x+7x=(5+7)x =?

Answer 17

(x+3)(x-3)

Answer 18

12x

Answer 19

yes, so numerator will be?

Answer 20

(x+3)(x-3)

Answer 21

that is denominator

Answer 22

yes so my awnser shoujld be \[\frac{ 12x-21 }{ (x+3)(x-3) }\]

Answer 23

that is absolutely correct.
if u want you can factor out 3 from numerator

Answer 24

ok thank you. soooo much

Answer 25

welcome ^_^

Answer 26

im gonna post a few more thanks for your help