# Math Help - [SOLVED] Rational Expressions Problem

1. ## [SOLVED] Rational Expressions Problem

Hi, i'm new the forums so pardon me if i posted this question in the wrong section. I'm having trouble with these problem. I think i've got the answer but whenever i substitue a value in for X in the simplified equation, i get a different answer than the value i got if i did so with the orginal equations. What i'm saying may not be making any sense to you right now so please check the picture of the question and attached.

THANKS.

2. Here is the problem: you are given that the area of a rectangle is given by $\frac{2x^2- 7x+ 6}{x^2- 1}$ and that the length of one side is $\frac{x-2}{x+1}$.

Since the area of a rectangle is given by the product of the two lengths, we find the length of the other side by dividing, and since we divide by a fraction by "inverting" and multiplying that is $\frac{2x^2- 7x+ 6}{x^2- 1}\frac{x+1}{x-2}= \frac{(2x- 3)(x-2)}{(x-1)(x+1)}\frac{x+1}{x-2}= \frac{2x-3}{x-1}$.
That you have right.

The perimeter of the rectangle is 2 times the sum of those two lengths. The sum of the lengths alone is $\frac{x-2}{x+1}+ \frac{2x-3}{x-1}$. To add we get the common denominator: multiply numerator and denominator of the first fraction by x-1 and the second fraction by x+1:
$\frac{(x-2)(x-1)}{(x+1)(x-1)}+ \frac{(2x-3)(x+1)}{(x+1)(x-1)}$ $= \frac{x^2- 3x+ 2+ 2x^2- x- 3}{x^2-1}= \frac{3x^2- 4x- 1}{x^2- 1}$

Now multiply that by 2: the perimeter is $\frac{6x^2- 8x- 2}{x^2- 1}$, exactly what you have.

What numbers did you try? If we take x= 3, the side given is (3-2)(3+1)= 1/4 and the area is (18- 21+ 6)/(9-1)= 3/8 so the other side is (3/8)(4/1)= 3/2. Then the perimeter is 3/2+ 1/4+ 3/2+ 1/4= (6+ 1+ 6+ 1)/4= 14/4= 7/2, while the formula you got gives (54- 24- 2)/(9- 1)= 28/8= 7/2, exactly the same.

3. Thanks so much for your help HallsofIvy. I think i know what i did wrong. I chose 1 to be X which doesn't work because it's a restriction. So i assume that choosing any number besides one that would be a restriction should indeed work. Thanks again for your help!