Question

@phi Question is below

Answer 1

can you help me with this I have no idea what to do

Answer 2

is the problem
\[ \frac{ \frac{4}{(x+3)}}{\frac{1}{x}+3}\]?

Answer 3

yes

Answer 4

I would first add
1/x + 3
to do that, use a common denominator of x. that means multiply the 3 (which is 3/1) by x/x
then add the tops
can you do that ?

Answer 5

I think so so multiply the bottoms of 3/1 and x/x then add 3+x ?

Answer 6

what do you get ?

Answer 7

3+x/x
right?

Answer 8

3 * x/x means 3/1 * x/x which means 3*x (top * top) divided by 1*x (bottom*botom)

in other words
\[ 3 \cdot \frac{x}{x} = \frac{3x}{x} \]
now add the fractions
\[ \frac{1}{x} + \frac{3x}{x} \]
what do you get ?

Answer 9

to add fractions with the same bottom, add the tops, and keep the bottom as it was.

Answer 10

1+3x/x
Right?

Answer 11

yes. so now your problem is
\[ \frac{ \frac{4}{(x+3)}}{\frac{3x+1}{x}}\]

we can now flip the bottom fraction, and multiply (instead of divide)
follow ?

Answer 12

what is the inverse of (3x+1)/x
in other words, can you write down the "flip" of that fraction ?

Answer 13

uhm you can make the bottom fraction x/(3x+1) which would make it 4/(x+3)*x/(3x+1) right?

Answer 14

you there?

Answer 15

@phi

Answer 16

yes, you are correct. you get
\[ \frac{4}{x+3} \cdot \frac{x}{3x+1} = \frac{4x}{(x+3)(3x+1)}

\]
which you may (depends on how the answers are shown) multiply out to
\[ \frac{4x}{3x^2+10x+3} \]