Question

u+3 divided by u²-9 restrictions on variables?

Answer 1

hint : \(\large u^2 - 9 = u^2 - 3^2\)

Answer 2

hint2 : \(\large a^2-b^2 = (a+b)(a-b)\)

Answer 3

i don't know? 3?

Answer 4

I already know the answer is 1 over u-3

Answer 5

i just need the restrictions on variables.

Answer 6

\[\large \dfrac{u+3}{u^2-9} = \dfrac{u+3}{(u+3)(u-3)}\]

Answer 7

what values of \(u\) make the denominator equal 0 ?

Answer 8

3 and -3?

Answer 9

Yes ! so the restrictions are \(u \ne \pm 3\)

Answer 10

The whole point @ganeshie8 is trying to make, in math we don't like the zero in the denominator, that is why this is a restriction :)

Answer 11

and as u said the expression itself simplifies to :
\[\large \dfrac{u+3}{u^2-9} = \dfrac{u+3}{(u+3)(u-3)} = \dfrac{1}{u-3}\]