Question

Simplify the complex fraction.

Answer 1

\[\Large\frac{ \frac{ x }{ x+3} }{ \frac{ 1 }{ x }+\frac{1}{x+3} }\]
@phi

Answer 2

is it the samething as last time?

Answer 3

I would multiply top and bottom by x(x+3)
as a first step

Answer 4

why?

Answer 5

notice if you multiply the first fraction in the bottom by x you "get rid" of the divide by x
ditto if you multiply the second fraction by (x+3)
so if we multiply the bottom by x(x+3) that will get rid of the fractions in the bottom
and to keep things equal, we have to do the same to the top: multiply by x(x+3)

Answer 6

if you think that is too mysterious, then you could
1) find the common denominator for the bottom fractions (it is x(x+3) )
and add the bottom fractions to get one fraction.
then invert and multiply

Answer 7

so, what exactly am I multiplying by x(x+3)

Answer 8

the bottom i.e. (1/x + 1/(x+3) )
and the top i.e. (x/x+3)
you will get a new bottom and a new top

Answer 9

confused?

Answer 10

I'm trying to do it, and it's not working out too well XD

Answer 11

here is just the bottom
\[ \left( \frac{ 1 }{ x }+\frac{1}{x+3}\right) x(x+3) \]

Answer 12

distribute the x(x+3)
\[ \left( \frac{ x(x+3) }{ x }+\frac{ x(x+3)}{x+3}\right) \]

Answer 13

okay, let me try the top

Answer 14

\(\Large\frac{x^2(x+3)}{x+3}\)

Answer 15

simplify
\[ \frac{ \cancel{x}(x+3) }{ \cancel{x} }+\frac{ x\cancel{(x+3)}}{\cancel{(x+3)}} \]
with \( x\ne 0, x \ne-3\)

Answer 16

so I have \(\Large\frac{x^2}{(x+3)+x}\), right?

Answer 17

yes, but notice you have x+x in the bottom

Answer 18

so \(\Large\frac{x^2}{2x+3}\). (I was getting there, haha)

Answer 19

yes, with the restrictions noted above, and the "obvious" restriction
i.e. we don't want 2x+3 = 0
which means
2x= -3
x= -3/2 is not allowed as a 3rd restriction

Answer 20

alright, thanks :)

Answer 21

yw

Answer 22

btw , if you added the two fractions using a common denominator of x(x+3) the work looks almost exactly the same.

Answer 23

oh, okay.

Answer 24

what you told me was simple enough :) so thank you

Answer 25

you would do this for the bottom
\[ \frac{ 1 }{ x } \frac{(x+3)}{(x+3)}+\frac{1}{(x+3)} \frac{x}{x} \\
\frac{x+3+x}{x(x+3)} \\
\frac{2x+3}{x(x+3)}
\]
then the problem is
\[ \frac{ \frac{ x }{ x+3} }{ \frac{ 1 }{ x }+\frac{1}{x+3} } \\
\frac{ \frac{ x }{ x+3} }{\frac{2x+3}{x(x+3)} }
\]
now "flip" the bottom mess and multiply:
\[ \frac{ x }{ x+3} \cdot \frac{x(x+3)}{2x+3}\]

Answer 26

your original explanation looks so much easier XD