Question

Would you help me finish solving this rational function?

Answer 1

\[\frac{ 2 }{ x+2 } +\frac{ 1 }{ x-2 } =\frac{ 13 }{ 21 }\]

Answer 2

I know my LCD needs to be 21(x+2)(x-2).

Answer 3

So when I multiply the left side of the equation the x+2 and x-1 cancel out and I am left with 21(2) and 21(1) so I get 42+21 on the left side.

On the right side 21 cancels out and I am left with 13(x+2)(x-2) so my right side of the equation looks like 13x+26+13x-26, simplify and then just 26x.

My equation now looks like 42+21=26x.
Did I do something wrong, I do not have an x^2(squared) value?

Answer 4

Does anyone have any ideas, or some sort of next step?

Answer 5

Are you looking for the solutions?

Answer 6

@iambatman I am asking whether or not the work I did so far is correct. I am doubting it because I did not end up with a quadratic equation.

Answer 7

I see, lets deal with the left side only first :)

Answer 8

@iambatman Ultimately I need to find the solutions but I need to know the steps to get there and am stuck thus far. Thank you so much!

Answer 9

\[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

Answer 10

@iambatman What is that called? Is it like the quadratic formula, I have not seen this yet in my class?

Answer 11

That's basically showing you how to do fractions, so, \[\frac{ 2 }{ x+2} + \frac{ 1 }{ x-2 } \implies \frac{ (x-2)(2)+(1)(x+2) }{ (x+2)(x-2) }\]

Answer 12

@iambatman Do you then simplify?

Answer 13

\[\frac{ 3x-2 }{ (x+2)(x-2) } = \frac{ 13 }{ 21 }\] you should end up with this, can you finish it off?

Answer 14

Cross multiply, yadi yada yada, and set it = 0 and you'll have a quadratic function! :)

Answer 15

You can use any method to find the solutions after, quadratic formula, complete the square, what ever!

Answer 16

@iambatman Okay, thank you for setting it up like that. Excuse my naive math skills. The way I was taught was to find the LCD which was 21(x+2)(x-2), multiply that on both sides, simplify then solve. I've never set it up like yours before.
If I'm multiplying 13(x+2)(x-2) would I multiply the (x+2)(x-2) first?

Answer 17

Yes :), you could also do it 21(x+2)(x-2), but eh, I think it's more tedious, the "formula" I presented to you is actually just how you do regular fractions :P, it's nothing new!

Answer 18

\[(3x-2)(21) = 13(x+2)(x-2)\]

Answer 19

I believe the quadratic may simplify pretty nicely actually, you might just have one big number, but it shouldn't be too bad, good luck!

Answer 20

@iambatman If I was to set it up using 21(x+2)(x-2) multiplying by both sides, on the left side the x+2 and x-2 would cancel out and I'd be left with 21(2) and 21(1)= 42 +21.
On the right side the 21 would cancel out and then I would have 13(x+2)(x-2).

So my equation would be:
21+42=13(x+2)(x-2)

Then
\[63=13(x^{2}-2x+2x-4)\]

Is that correct up until this point?

Answer 21

Yes, that looks good :)

Answer 22

Wait, hold on

Answer 23

@iambatman When I simplified the right side I got 13x^2 - 52, still no x value?

Answer 24

No, that's not right, 21(x+2)(x-2) you should get 21x^2-84

Answer 25

@iambatman So I need to simplify the 21(x+2)(x-2) before I multiply that LCD on both sides of the equation?

Answer 26

Ok let me see what you're trying to do, mhm.

Answer 27

\[\frac{ 3x-2 }{ (x+2)(x-2) } - \frac{ 13 }{ 21 } = 0 \implies \frac{ 21(3x-2)-13(x+2)(x-2) }{ 21(x+2)(x-2) }\] this is what you get if you want the 21 in your denominator.

Answer 28

= 0 of course

Answer 29

That's what I'm getting with what you're trying to do, but it's not very clear to me, maybe if you can draw it out/ use LaTeX.

Answer 30

@iambatman I so appreciate you helping me out. \[21(x+2)(x-2) \times (\frac{ 2 }{ x+2 } + \frac{ 1 }{ x-2 }) = (\frac{ 13 }{ 21 }) \times 21(x+2)(x-2)\]

Answer 31

@iambatman Does this help make it more clear?

Answer 32

That should work also :)

Answer 33

It will simplify to \[63x-42=13(x^2-4)\] which will give you the same result

Answer 34

@iambatman I understand how you got the right side of the equation but could you type out how you got the left side of the equation?
The 63x -42?