I'm so lost on this equation, please help.

[Scotty_Smith]

I'm taking a college course where the instructor doesn't teach anything. It's not an online course, but we do all the learning online. We had a test last week and I did pretty bad, the problems that are giving me the most trouble are:

Solve the equation and check for extraneous solutions.
(9/(x+2)) -3=((x+11)/(x+2))

she crossed out all of my work and wrote the answer as: x=6

the second problem is:

Solve the equation and check for extraneous solutions.
((x^2-20)/(x^2-7x+12))=(3/(x-3))+(5/(x-4))

for this one, the answer was: x=1,7

I've been trying to solve these since Thursday and I'm not getting any where. Thank you for any help!

 

[DrPhil]

Solve the equation and check for extraneous solutions.
(9/(x+2)) -3=((x+11)/(x+2))

she crossed out all of my work and wrote the answer as: x=6

the second problem is:

Solve the equation and check for extraneous solutions.
((x^2-20)/(x^2-7x+12))=(3/(x-3))+(5/(x-4))

for this one, the answer was: x=1,7

Show us your work so we can see where you are getting stuck.

When dealing with rational equations, you can generally say that the denominator tells you what values are NOT allowed, and the numerator tells you where the solutions are.

Problem 1, first thing to note is that x may not be -2, because (x+2) occurs in denominators. Then you may multiply all terms in the equation by (x+2):
9 - 3(x + 2) = x + 11
. . .

For the 2nd problem, the LCD is (x-3)(x-4), so values x=3 or x=4 are not allowed. Can you place all terms over the LCD, and then solve the quadratic equation in the numerator?

 

[Scotty_Smith]

I know the answer isn't x=6 Dennis. That's why I'm confused and that's why I'm here. Everything in my post is accurate to what is on my paper.

So far what I keep doing is multiplying the equation by x+2 so that I can get rid of fractions. so the equation becomes:
9-3(x+2)=x+11.
Distribute:
9-3x-6=x+11
combine like terms
3-3x=x+11
x on one side
-4x=8
divide
x=-2

But when I plug that into the original problem it doesn't work. I end up with
(9/0)-3=(9/0)

 

[JeffM]

I'm taking a college course where the instructor doesn't teach anything. It's not an online course, but we do all the learning online. We had a test last week and I did pretty bad, the problems that are giving me the most trouble are:

Solve the equation and check for extraneous solutions. There is something very odd about this problem. Extraneous solutions? There is no solution to the equation you show below, namely

\(\displaystyle x \ne - 2\ and\ \dfrac{9}{x + 2} - 3 = \dfrac{x + 11}{x + 2} \implies 9 - 3(x + 2) = x + 11 \implies 9 - 6 - 11 = x + 3x \implies x = - \dfrac{8}{4} = - 2.\) Contradiction.

(9/(x+2)) -3=((x+11)/(x+2)) Was this the original problem?

she crossed out all of my work and wrote the answer as: x=6

the second problem is:

Solve the equation and check for extraneous solutions.
((x^2-20)/(x^2-7x+12))=(3/(x-3))+(5/(x-4))

for this one, the answer was: x=1,7 This is the correct answer as you can tell by inserting the values of 1 and 7 into the equation. Please show your work so we can see where you have gone astray.

I've been trying to solve these since Thursday and I'm not getting any where. Thank you for any help!

.

 

[Scotty_Smith]

Yes, the first one is the original problem. I was able to complete the second one by multiplying the LCD and solving normally getting x^2-8x+7=0 factoring that into (x-7)(x-1)=0. x=7, x=1.
Every one I ask tells me the same thing; there's no solution for the 1st equation. I'll just have to ask my instructor to prove her answer on Wednesday. >_>
Thanks for all of your answers on this question.