# Hi guys....I'm having the toughest time with one problem!

• Jul 27th 2009, 07:08 PM
byco
Hi guys....I'm having the toughest time with one problem!
Hi, I'm Brad and new here.

I'm having the hardest time with this one problem in intermediate algebra. It wants me to solve for X but every time I try to solve the problem I get about half way through and get stuck. I think this one problem is literally taking years off my life. Any help would be GREAT!!!!

http://www.screencast.com/users/Bste...5-6c4010f76e2d

2009-07-27_2152

I posted an image with the problem but if that doesn't work, the link takes you to screencast where I wrote out the problem.

Thanks Guys!!
• Jul 27th 2009, 07:18 PM
artvandalay11
$\displaystyle \frac{x+5}{x^2+3x+2}-\frac{5}{x^2+4x+4}=\frac{x-5}{x^2+3x+2}$

$\displaystyle \frac{x+5}{x^2+3x+2}-\frac{x-5}{x^2+3x+2}=\frac{5}{x^2+4x+4}$

$\displaystyle \frac{x+5-x+5}{x^2+3x+2}=\frac{5}{x^2+4x+4}$

$\displaystyle \frac{10}{x^2+3x+2}=\frac{5}{x^2+4x+4}$

Now we will cross multiply to get

$\displaystyle 10(x^2+4x+4)=5(x^2+3x+2)$ and divide by 5 to get

$\displaystyle 2(x^2+4x+4)=x^2+3x+2$

$\displaystyle 2x^2+8x+8=x^2+3x+2$ and bring it all over to the left, factor what's left and you should be able to get the answer

Be sure to check both answers by plugging back in though as one of the solutions will lead to a zero in the denominator, which means it must be thrown out
• Jul 27th 2009, 07:20 PM
VonNemo19
Quote:

Originally Posted by byco
Hi, I'm Brad and new here.

I'm having the hardest time with this one problem in intermediate algebra. It wants me to solve for X but every time I try to solve the problem I get about half way through and get stuck. I think this one problem is literally taking years off my life. Any help would be GREAT!!!!

http://www.screencast.com/users/Bste...5-6c4010f76e2d

2009-07-27_2152

I posted an image with the problem but if that doesn't work, the link takes you to screencast where I wrote out the problem.

Thanks Guys!!

Subtract both sides: $\displaystyle \frac{x+5}{x^2+3x+2}-\frac{x-5}{x^2+3x+2}-\frac{5}{x^2+4x+4}=0$

Note the common denominator allows me to subtract the first two terms.

$\displaystyle \frac{10}{x^2+3x+2}-\frac{5}{x^2+4x+4}=0$

Now we've gotta find a common denominator.

$\displaystyle \frac{10}{(x+1)(x+2)}-\frac{5}{(x+2)^2}=0$

Then
$\displaystyle \frac{10(x+2)}{(x+1)(x+2)^2}-\frac{5(x+1)}{(x+1)(x+2)^2}=0$

$\displaystyle \frac{10(x+2)-5(x+1)}{(x+1)(x+2)^2}=0$

Multiply both sides by the denominator getting

$\displaystyle 10(x+2)-5(x+1)=0$

Surely you can finish up!(Wink)
• Jul 27th 2009, 07:25 PM
ANDS!
The thing with "big ugly" problems, is that there is usually something that can be immediately done to a problem to make it not so big and ugly. I see one thing right off the bat.

There is a factor of (X+2) in all of those equations. Once you do that, it should become clear what you must do next.
• Jul 27th 2009, 07:25 PM
Rapha

Quote:

Originally Posted by byco
Hi, I'm Brad and new here.

I'm having the hardest time with this one problem in intermediate algebra. It wants me to solve for X but every time I try to solve the problem I get about half way through and get stuck. I think this one problem is literally taking years off my life. Any help would be GREAT!!!!

http://www.screencast.com/users/Bste...5-6c4010f76e2d

2009-07-27_2152

I posted an image with the problem but if that doesn't work, the link takes you to screencast where I wrote out the problem.

Thanks Guys!!

$\displaystyle \frac{x+5}{x^2+3x+2}-\frac{5}{x^2+4x+4} = \frac{x-5}{x^2+3x+2}$

Minus $\displaystyle \frac{x+5}{x^2+3x+2}$ leads to

$\displaystyle -\frac{5}{x^2+4x+4} = \frac{x-5}{x^2+3x+2} -\frac{x+5}{x^2+3x+2}$

$\displaystyle -\frac{5}{x^2+4x+4} = \frac{(x-5)-(x+5)}{x^2+3x+2}$

$\displaystyle -\frac{5}{x^2+4x+4} = \frac{-10}{x^2+3x+2}$

Multiply by x^2+4x+4

$\displaystyle -5 = \frac{-10(x^2+4x+4)}{x^2+3x+2}$

Multiply by x^2+3x+2

$\displaystyle -5(x^2+3x+2) = -10(x^2+4x+4)$

Do you know how to solve this?

There are two solutions, x = -3 ; x = -2

Yours
Rapha

Edit: Beaten to it by ...
• Jul 27th 2009, 07:25 PM
Krahl
that can be simplified to...

(x+5)/((x+2)(x+1)-5/((x+2)^2)=(x-5)/((x+2)(x+1))

multiply through by x+2 since x can't be -2...

(x+5)/(x+1)-5/(x+2)=(x-5)/(x+1)

rearrange to get...

(x+5)/(x+1)-(x-5)/(x+1)=5/(x+2)

then

((x+5)-(x-5))/(x+1)=5/(x+2)

10/(x+1)=5/(x+2)

then solve to get x=-3 and check that it satisfies the equation

is -2 a solution Rapha? it makes the denominator 0 but not the numerator...
• Jul 27th 2009, 07:29 PM
VonNemo19
Quote:

Originally Posted by Rapha
There are two solutions, x = -3 ; x = -2
Edit: Beaten to it by ...

But, surely there is only one! It's all in the domain my dear friend!
• Jul 27th 2009, 07:31 PM
Rapha
Quote:

Originally Posted by Krahl

is -2 a solution Rapha? it makes the denominator 0 but not the numerator...

-2 is a solution of the equation
http://www.mathhelpforum.com/math-he...ff658d4f-1.gif

but not a solution of the original post, because ... (you noticed the thing about the denominator (Wink) )
• Jul 27th 2009, 07:33 PM
Krahl
can anyone spot where rapha went wrong lol
• Jul 27th 2009, 07:36 PM
VonNemo19
Quote:

Originally Posted by Krahl
can anyone spot where rapha went wrong lol

Rapha did not "go wrong". Note that he has obtained the solution, but $\displaystyle x=-2$ is not in the Domain of the equation, therefore it is what we call an "extraneous" solution.
• Jul 27th 2009, 07:53 PM
byco
Man, you guys made this look so easy.(Clapping)

Seriously though, thank you all. You guys really helped me understand it.

Thanks!!
• Jul 27th 2009, 07:59 PM
VonNemo19
Quote:

Originally Posted by byco
Man, you guys made this look so easy.(Clapping)

Seriously though, thank you all. You guys really helped me understand it.

Thanks!!

That's what we're here for buddy.(Cool) And thank you for being grateful. Some don't even bother...
• Jul 28th 2009, 06:48 AM
pacman
I agree, x = -3 is the only solution.
• Jul 28th 2009, 07:40 AM
pacman
see the figure