Rational expression

G

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Perform the indicated operation
1/a-2/a-1+3/a-1

1(a-1)(a+1) - 2(a)(a-1) +3(a)(a+1)
a(a-1)(a+1) a+1(a)(a-1) a-1(a)(a+1)

(2a^2)-2a(2a-2)+3a(3a+3)

2a^2-a(5a+1)
a(a-1)(a+1)

The a in front of the second term is bothering me I don't think it is suppose to be there, but I don't know where I went wrong.
 
Your formatting (specifically, the lack of grouping symbols) is ambiguous, but I will guess from your working that you mean the following:

. . . . .\(\displaystyle \displaystyle \large{\frac{1}{a}\mbox{ } - \mbox{ }\frac{2}{a - 1}\mbox{ } +\mbox{ } \frac{3}{a - 1}}\)

However, I am not sure where you are getting an "a + 1" factor in the denominators...?

Please reply with clarification. Thank you.

Eliz.
 
bethany3168 said:
Perform the indicated operation
1/a-2/a-1+3/a-1

1(a-1)(a+1) - 2(a)(a-1) +3(a)(a+1)
a(a-1)(a+1) a+1(a)(a-1) a-1(a)(a+1)

(2a^2)-2a(2a-2)+3a(3a+3)

2a^2-a(5a+1)
a(a-1)(a+1)

The a in front of the second term is bothering me I don't think it is suppose to be there, but I don't know where I went wrong.
Click to expand...

BRACKETS SVP!! The way you're showing it means:
1/a - 2/a - 1 + 3/a -1
=1/a - 2/a + 3/a - 2
= 2/a - 2

Remember that (as example) 2/a-1 means 2/a less 1, not 2 divided by a-1;
here's a clear example:
24/4+2 = 6 + 2 = 8
24/(4+2) = 24/6 = 4 : what's in brackets is always done first

Same thing applies to multiplications and divisions;
20/2*5 = 10 * 5 = 50
20/(2*5) = 20/10 = 2

If what you mean is (as stapel is showing):
1/a - 2/(a-1) + 3/(a-1), then you can first simplify to:
1/a + 1/(a-1)
then you need to multiply each term by the lcm and you'll get:
(a - 1 + a) / [a(a - 1)]
= (2a - 1) / [a(a - 1)]
 
My example is set up like Stapel posted, however under the 2 it should be a+1

Our answer choices are as follows:

2a^2+5a+1
a(a-1)(a+1)

2a^2+5a-1
a(a-1)(a+1)

2a^2+5a-1
a(a-1)^2

I have been trying to work the problems first and then look at the answer choices. I am close, but I had that extra a and I am not sure why.
 
bethany3168 said:
My example is set up like Stapel posted, however under the 2 it should be a+1....I have been trying to work the problems...I am close, but I had that extra a and I am not sure why.
Click to expand...
We will be glad to check your work, but you'll need to post it first.

Thank you.

Eliz.
 
I posted all my work in the first post, as you can see my answer does not match the choices given...

1(a-1)(a+1) - 2(a)(a-1) +3(a)(a+1)
a(a-1)(a+1) a+1(a)(a-1) a-1(a)(a+1)

(2a^2)-2a(2a-2)+3a(3a+3)

2a^2-a(5a+1) I have an extra a in front of the 5??
a(a-1)(a+1)

Our answer choices are as follows:

2a^2+5a+1
a(a-1)(a+1)

2a^2+5a-1
a(a-1)(a+1)

2a^2+5a-1
a(a-1)^2
 
As has been mentioned a few times in the past, when one doesn't use grouping symbols, one's meaning can be very difficult to ascertain.

I'm afraid I can't follow what you've posted. Sorry.

Eliz.
 
The 2nd choice is the answer:
2a^2+5a-1
a(a-1)(a+1)
That's a division; should be shown as:
2a^2+5a-1
--------------
a(a-1)(a+1)

or as (2a^2+5a-1) / [a(a-1)(a+1)]

So your original expression (properly bracketed!) is:
1/a - 2/(a+1) + 3/(a-1)

Applying lcd of a(a+1)(a-1) gives:
[(a+1)(a-1) - 2(a)(a-1) + 3(a)(a+1)] / [a(a-1)(a+1)]
= (a^2 - 1 - 2a^2 + 2a + 3a^2 + 3a) / [a(a-1)(a+1)]
= (2a^2+5a-1) / [a(a-1)(a+1)]
 
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