# Easy Rational Expression problems which I've completed and seeking review from peers

• Mar 5th 2013, 06:57 PM
Odonsky
Easy Rational Expression problems which I've completed and seeking review from peers
Hi all,

Below are several problems I believe I have set up and solved correctly but would like for others to review them as well.

1). x / x^2 - 3x - 18 multiplied by x^2 - x - 12 / x^2 - 4x

= x / (x - 6) (x + 3) multiplied by (x - 4) (x + 3) / x (x - 4)

= 1 / x - 6 <-- I never factored in a LCD to get this answer, simply reduced.

2). x^2 - 3x - 4 / x^2 - 16 multiplied by x +4 / 3x - 9

= (x - 4) (x + 1) / (x + 4) (x - 4) multiplied by x + 4 / 3 (x - 3)

= x + 1 / 3 (x - 3) <-- Again, no LCD was factored into the equation, simple reduction only.

3). x - 3 / x + 2 divided by x^2 - x - 6

= x - 3 / x + 2 multiplied by 1 / (x - 3) (x + 2)

= 1 / (x + 2) (x + 2) or 1 / (x + 2)^2

4). 4 / x + 3 subtracted by 6 / 3x + 9

= 4 / x + 3 subtracted by 6 / 3 (x + 3); LCD equals 3 (x + 3)

= 3 (x + 3) times 4 / x + 3 subtracted by 6 / 3 (x + 3)

= 12 - 6 / 3 (x + 3) or 2 / x +3 when simplified

5). x + 3 / x^2 - x - 12 plus 2 / x +3 = 1 / x - 4

= x + 3 / (x - 4) (x + 3) plus 2 / x + 3 = 1 / x - 4; LCD = (x - 4) (x + 3)

= (x - 4) 2 = x + 3
= 2x - 8 = x + 3
2x = x + 11
x = 11

Is anyone able to verify that my equations are set up properly and with correct answers?

Thank you.
• Mar 5th 2013, 08:06 PM
ibdutt
Re: Easy Rational Expression problems which I've completed and seeking review from pe
The first question second rational expression has been wrongly factorized. It should be (x+3)(x-4)/x(x-4) Now when you multiply you will get the correct answer i.e., 1/(x-6).
In second question answer use parenthesis to avoid confusion: (x+1)/[3(x-3)]
Third is correct. In fourth use brackets.
Fifth question is not clear
• Mar 5th 2013, 09:05 PM
Odonsky
Re: Easy Rational Expression problems which I've completed and seeking review from pe
Quote:

Originally Posted by ibdutt
The first question second rational expression has been wrongly factorized. It should be (x+3)(x-4)/x(x-4) Now when you multiply you will get the correct answer i.e., 1/(x-6).
In second question answer use parenthesis to avoid confusion: (x+1)/[3(x-3)]
Third is correct. In fourth use brackets.
Fifth question is not clear

Hi,

You are correct, the first question was not factored properly and I have since adjusted it.

The fifth question is meant to be solved. I have rewritten the equation, hopefully it is easier to read.

(x + 3) / [x^2 - x - 12] + 2 / (x + 3) = 1 / (x - 4).
• Mar 5th 2013, 10:20 PM
ibdutt
Re: Easy Rational Expression problems which I've completed and seeking review from pe
Still it does not make sense. If we take it as it is then we have
(x+3)/[(x+3)(x-4) ] + 2/(x-3) = 1 / (x-4)
1/(x-4) + 2/(x-3) = 1 / (x-4)
2/(x-3) = 0 That just indicates that x is infinitely large
• Mar 6th 2013, 06:24 PM
Odonsky
Re: Easy Rational Expression problems which I've completed and seeking review from pe
Quote:

Originally Posted by ibdutt
Still it does not make sense. If we take it as it is then we have
(x+3)/[(x+3)(x-4) ] + 2/(x-3) = 1 / (x-4)
1/(x-4) + 2/(x-3) = 1 / (x-4)
2/(x-3) = 0 That just indicates that x is infinitely large

After factoring and reducing the equation aren't we left with the following:

x + 3 + 2x - 8 = x + 3

Combine like terms:

3x - 5 = x + 3

Subtract x from both sides:

2x - 5 = 3

2x = 8

Divide 2 to both sides:

x = 4

This answer is different than the original one I posted because I have since re-worked the problem.

However, when I plug 4 into all x values I end up with denominators equal to 0 which makes the equation 'invalid.'

Determining whether the equation has a solution or is invalid was part of the instructions so does x = 4 sound correct?
• Mar 6th 2013, 06:34 PM
Odonsky
Re: Easy Rational Expression problems which I've completed and seeking review from pe
When is it necessary to arrange all terms and set to standard form?