Question

I know that the solution is x=7, but how do I solve for the extraneous solution (which I now know is x=-2)?

Answer 1

do you mind showing me your work and how you got one of the answer, 7? and maybe we can work and see whats wrong

Answer 2

I think I've figured it out, but I'll show you. Maybe you can tell me if I'm doing this right.

Answer 3

First you need to get all the denominators to match up.
3/x^2+5x+6 has the LCD.

To make the other fractions have the same denominator, multiply "x-1/x+2" by "x+3" and 7/x+3 by "x+2".
x-1/x+2(x+3)=x^2+2x-3
7.x+3(x+2)=7x+14

Now set up the numerators as an equation:
3+x^2+2x-3=7x+14

Combine like terms:
x^2+2x=7x+14

Subtract 2x from both sides:
x^2=5x+14

This is where I get stuck. I know you have to try and factor the equation.
(x-2)(x+7) is needed to make 5x, but (x+2)(x+7) is needed to make 14. Because 7 is in both versions, I chose it as the solution, but how do I know which is the extraneous solution between -2 and 2?

Answer 4

Please let me if my wording is confusing.

Answer 5

you got the right concept. but can i help you clarify you steps better? what your're doing is right but you r steps arent really that right.

Answer 6

Yeah I've forgotten some of the official terms.

Answer 7

your first step about finding the LCD is correct.

however, when you multiply, you do not want to just multiply by x+3 or just x+2
you want to multiply it like this
\[\frac{ 3 }{ (x+2)(x+3) }+\frac{ x-1 }{ x+2 }\left( \frac{ x+3 }{ x+3 } \right)= \frac{ 7 }{ x+3 }\left( \frac{ x+2 }{ x+2 } \right)\]

Answer 8

becuz x+3/x+3 and x+2/x+2 is like multiply those fractions by 1

Answer 9

and while we here.
you see how all the fractions have a denominator of (x+2)(x+3) and remember how denominators cant equal 0.
we set up this equation
(x+2)(x+3)=0
and we get x=-2 and -3
and you should write them somewhere that x CANNOT equal -2 or -3

are you with me so far?

Answer 10

The first part I do generally right out. I guess I was just trying to shorten up my response. I apologize for the confusion.
And yes I am.

Answer 11

Then multiply both sides of the equation by (x+2)(x+3) so you're just left with ur numerator
And everything's good till you get to x^2 =5x+14

Your next step would be to make the equation equal to zero so you can factor.
So
x^2 =5x+14
-5x. -5x
---------------
x^2 -5x=14
-14. -14
-------------
x^2 -5x-14=0

So can you factor this equation?

Answer 12

Yes. (x+2)(x-7)

Answer 13

yes. so (x+2)(x-7)=0
so that means either one of the paranthesis must equal zero
so set up these 2 equations

x+2=0 and x-7=0
what are the two possible solutions for x?

Answer 14

x=-2 and x=+7
and since -2 would cause the denominator to equal 0 in the original equations, that why it's the extraneous solution?

Answer 15

correct

Answer 16

Ah. It all makes sense now. Thank you very much. :)

Answer 17

yea no problem.

i think your name problem was that you didnt find the answers that wouldnt work first
and that you forgot to set the equation to equal 0 before you factor

Answer 18

Yeah I knew that I there was something I wasn't doing right. I'm all set now though. :)