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June 6th, 2014, 03:09 PM   #1
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Could someone please explain these question?

I do not know how to do these but I tried. These are rational expressions. Please explain them, I want to learn how to do them. Thank you.

1. 1/x−1 + 4/x²−1 − 2/ x²−2x+1

my answer 1/x-1 + 4/ (x-1) (x+1) - 2/ (x-1) (x-1)


2. 9 -1/y²/ 3- 1/y

my answer: 6y

3. 5/t-1 + 3/t

my ans 8/ t(t-1)
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June 6th, 2014, 04:06 PM   #2
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The aim is to merge all the terms in one quotient. We do this by finding a denominator that is the product of all the distinct elements in the denominators of the individual terms. The easiest way to do this is to simply multiply the denominators, but that is often more than you need to do and it can make the next part more complicated.

Let's look at the first question.

There are three denominators here, but your work in factorising then (a good first step) has shown that each contains $(x - 1)$, so we'll need that in the combined denominator. Looking at the second term, we see an $(x + 1)$ in addition to the $(x - 1)$, so we'll need that too. And the third denominator has a second $(x - 1)$ term, so we'll include that too. Thus the denominator of our answer will be $(x - 1)(x + 1)(x - 1) = (x - 1)^2(x + 1)$.

Now we need to construct our numerator. Starting with the first term, we see in going from the original denominator $(x - 1)$ to the combined denominator $(x -1)^2(x+1)$ we have multiplied by $(x-1)(x+1)$. With any fraction, if we want it to retain the same value, we must multiply the top by the same value as we did the bottom. In this case, we get $(x - 1)(x +1)$. This is the first term in our numerator.

The denominator of the second term has been multiplied by $(x-1)$, so we'll do the same to the numerator to get $4(x-1)$. Finally, the third denominator was multiplied by $(x+1)$, so we do that to the numerator getting $2(x+1)$.

Now we have scaled the numerators to preserve the value of the individual terms while giving them a common denominator. So we are ready to combine them, taking care to copy the operators from the original expression (in this case + then -).

So $$\frac{1}{x-1} + \frac{4}{(x-1)(x+1)} - \frac{2}{(x-1)(x+1)} = \frac{(x-1)(x+1)}{(x-1)^2(x+1)} + \frac{4(x-1)}{(x-1)^2(x+1)} - \frac{2(x+1)}{(x-1)^2(x+1)} = \frac{(x-1)(x+1) + 4(x-1) - 2(x+1)}{(x-1)^2(x+1)}$$
Now we have a single expression as required. The numerator needs to be simplified, but that is all.

You try the others again.

Last edited by skipjack; June 7th, 2014 at 03:07 AM.
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June 6th, 2014, 08:21 PM   #3
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They all boil down to this:
a/b + c/d = (ad + bc) / (bd)

1/2 + 1/4 = 2/4 + 1/4 = (2+1) / 4 = 3/4
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June 7th, 2014, 04:47 PM   #4
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Than you. I have figure how to do these.
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