Question

solve
1/(x+2) + 2/(x+4) = 0

Answer 1

\[y= \frac{ 1 }{ x+2 } + \frac{ 2 }{ x+4 } \] when y = 0

Answer 2

Is this a y= or an =0?

Answer 3

Ah, I see it now. You put it below.

Answer 4

OK, so, where are you getting stuck?

Answer 5

i don't understand if i should multiply the denominator or not. I guess i'm just stuck on how to start it

Answer 6

what i meant is, should i add them first and then get x by itself or should i do it separately

Answer 7

First, you have a rewstricted domain. Do you know what the restrictions are and why? It is important to note that up front.

Answer 8

...restricted domain.

Answer 9

yes i know what those are, but i don't really understand them

Answer 10

OK. Well, is \(\dfrac{1}{0}\) a valid fraction?

Answer 11

no its not

Answer 12

Exactly. That is why you have a restricted domain.

In the Cartesian plane, \(\sqrt{-1}\) is also impossible. That is a different type of restricted domain.

In short, division by 0 and negative roots are domain restrictions. That is all you need to understand to find out when there is a restriction.

Answer 13

Well, - even roots. \(\sqrt[3]{-1}\) does have a solution...

Answer 14

basically anything that is imginary or can't find the answer to is restricted, correct?

Answer 15

Exactly!

Answer 16

So, because you have \(\dfrac{ 1 }{ x+2 }\) and \(\dfrac{ 2 }{ x+4 }\), your restrictions are: \(x+2 \ne 0\) and \(x+4\ne 0\). Those solve very easily to \(x\ne \{-2 , -4\}\)

Answer 17

OK, with that in mind, your comment of "should i add them first and then get x by itself " is exactly correct.

Answer 18

So you need a common denomuinatior, do addition, then you can get everything over to one side, factor, and have somethign to work with.

Answer 19

so now i have \[\frac{ 3x+8 }{ x ^{2}-6x+8 }\]

Answer 20

And in this case, I do not see much factoring to do.

Answer 21

when working with fraction first find the lowest common denominator then multiply each term by this . since they will all have the same denominator you can just work with the numerator.

Answer 22

\(\dfrac{ 3x+8 }{ x ^{2}-6x+8 }=0\)

Yes, that is good. Now, if you multiply through to get rid of the fraction, what happens?

Answer 23

it becomes 3x+8 = 0 and then simply dived you get -8/3

Answer 24

omg thank you!

Answer 25

Yes. And is that part of the restrictions?

Answer 26

no?

Answer 27

Correct, so it is a valid answer.

And here is a way to confirm it:
https://www.desmos.com/calculator/rs3ucl7rgp

Answer 28

Let me add the restrictions to that and you qwill see why they are important.

Answer 29

Here we are:
https://www.desmos.com/calculator/zh11rbxpzf

See the blue dashed lines at x=-2 and x=-4? Those are asymptotes. The graph of your original equation never touches them.

Answer 30

yes i see that

Answer 31

are the restrictions always the asymptotes?

Answer 32

I hope that makes restrictions a little more clear. That is one thing that restrictions are. Places on a graph that are never touched.

Answer 33

A removable restriction can be a hole, which is different from an asymptote, but stil does not exist.

Answer 34

right, holes i understand

Answer 35

Here is the simple example: \(\dfrac{(x+1)(x-1)}{x+1}\)

That graphs as a line. Bit really, it is a line with a hole at x=-1.

Answer 36

So on a graph, restrictions are places that don't exist. Which makes sense with what you said earlier: They are things you can't solve. Well, solving means a y value. No y value and they don't exist. It is all related. A graph is visually and solving it is in numbers. Just two ways of saying the same thing. I just like graphs because visuals are powerful memory tools.

Answer 37

I know all that is a little more than answering your question, but I hope this makes any other ones you have to work on easier because the underlying concepts will be more clear.

=)

Answer 38

yes it did actually helped me completed the rest of my hw. thanks for the help i appriciate you taking the time to actually teach me

Answer 39

np. Have fun!

Answer 40

That was an awesome explanation!