Question

(4/x) + (4/x^2 - 9) = (3/x - 3)

Answer 1

@absurdism

Answer 2

Okay...I've worked on this one for ages. I think the LCD is x(x + 3)(x - 3). Want me to type up my work?

Answer 3

let's just go through it

Answer 4

K. Following your lead.

Answer 5

to make this easy i'm just gonna multiply everything by x^2 to get rid of those ugly variables in the denominator

4x + 4 - 9x^2 = 3x - 3x^2
-6x^2 + x + 4 = 0
can you factor that

Answer 6

I have another almost just like this, so when you've walked me through this one maybe I'll be able to solve the second.

Answer 7

Um..I'm not sure I know how to, because I don't know of any factors of -24 that sum to 1. Show me?

Answer 8

Also, how did you get x^2?

Answer 9

Or rather, know to use x^2.

Answer 10

I don't mean to make you do my work for me, I just don't know where to start there. @absurdism

Answer 11

well it was just logic
i have a weird intuition for problems like this (for all of math really), so i do things in a weird way but it works
if we multiply everything by x^2 then it gets rid of that one x^2 in the denominator and the other x's in the other denominators
idk if you understand but it makes sense to me

Answer 12

I get why it cancels the x^2 in the middle denominator, but I guess I thought it would create x^3 in the other two. I don't think I could repeat that method.

Answer 13

nope, 4/x * x^2 = (4x^2)/x = 4 * x^2/x = 4x

Answer 14

ew that looks so gross one sec

Answer 15

$$\frac{4}{x}(x^2) = \frac{4x^2}{x} = 4 \frac{x^2}{x} = 4x$$

Answer 16

*cries* @absurdism

Answer 17

\[\dfrac{4}{x}+\dfrac{4}{x^2}-9=\dfrac{3}{x}-3\]
Is this your problem??

Answer 18

ya i could have the problem wrong

Answer 19

I have to confirm first.

Answer 20

No, Bebong. Absurdism was right. I had them in parentheses to avoid it looking like that.

Answer 21

I get it now, absurdism. K...on to the next step?

Answer 22

\[\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}\]

Answer 23

Yes, that is correct @Bebong

Answer 24

So for the next part, do we get 4/-9??

Answer 25

I would clear the denominators.
Multiply both sides by the LCD x(x-3)(x+3)
\[ \dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3} \\
4(x-3)(x+3) + 4x = 3x(x+3)
\]
I skipped some steps, but if you multiply each term by the LCD and simplify you get that result.
Now expand
\[ 4(x^2 -9) + 4x = 3x^2 +9x\\
4x^2 -36 +4x = 3x^2 +9x \\
x^2 -5x -36=0 \]

Answer 26

I get the right side of that...how did you keep 4(x^2 - 9) + 4x from becoming 4(x^2 - 9 + x)???

Answer 27

do you see how I got the second line
\[ 4(x-3)(x+3) + 4x = 3x(x+3) \]

Answer 28

Whoohoo, that gives me a usable answer...it factors out to (x - 9)(x + 4), which gives me the solutions of x = 9, x = -4.

Answer 29

Yes, I get the factoring - what I don't get is how you kept the individual x from the LCD out of the parentheses...

Answer 30

Because when I multiply 4 by x(x - 3)(x + 3), I get 4(x^2 + x - 9). And that doesn't factor.

Answer 31

ok, in slow motion
\[ \cancel{x}(x - 3)(x + 3)\cdot \dfrac{4}{\cancel{x}}+x\cancel{(x - 3)(x + 3)}\cdot \dfrac{4}{\cancel{x^2-9}}=\dfrac{3}{\cancel{x-3}} \cdot x\cancel{(x - 3)}(x + 3) \]
that gives
\[ 4(x-3)(x+3) +4x = 3x(x+3) \]
the first term is 4(x^2-9) = 4x^2-36

Answer 32

**Because when I multiply 4 by x(x - 3)(x + 3), I get 4(x^2 + x - 9)
remember the first term is 4/x so the x's cancel

Answer 33

AHA!

Answer 34

it's not a coincidence the denominators "go away". That is what happens when we multiply by the LCD

Answer 35

I get it!! Thank you @absurdism @phi you guys rock :)

Answer 36

Sorry I can't give you both medals. @phi can you help me with one more like this - mainly just watch to make sure I'm doing the steps right like you showed me?

Answer 37

please make it a new post.

Answer 38

Sure :) Thanks!

Answer 39

sorry i couldn't explain as well as phi could. ^-^

Answer 40

That's fine @absurdism you still helped a lot!