How do you add rational expressions with different denominators and numerators?

Question

The problem is :

(3x+3/x^2+2x+1)+(x-1/x^2-1)

^2=squared

ive been on a lot of math help websites but I cant seem to figure out how to solve this

 

Answer 1

Daisy:

It seems like your other responders are using a different starting expression than I am reading in your post (?).  Correct me if I'm wrong, but the expression you have written (I believe) is as follows:

(3x+3/x²+2x+1)+(x-1/x²-1)

I would start by removing the parentheses and combining like terms to get 3x+2x+x+3/x²-1/x²+1-1

Next simplify to get 6x+2/x²

You can then multiply the first term by x²/x² to get the two remaining terms over the same denominator: 6x³/x²+2/x²

Now add the numerators to get the final result of (6x³+2)/x²

Again, if my read on your expression is correct!

George T.

Answer 2

Simplify first by factoring,

(3x+3)/(x^2+2x+1) + (x-1)/(x^2-1)

= 3(x+1)/(x+1)^2 + (x-1)/[(x+1)(x-1)]

= 3/(x+1) + 1/(x+1)

= 4/(x+1), where x ≠ 1.

Answer 3

Just as with adding fractions such as 1/4 and 1/3, we must have a common denominator. The least common multiple of 3 and 4 is 12.  You need the least common multiple of x^2 + 2x + 1 and x^2 - 1.  To do so we must first see what common factors they have, and in order to do so we must factor each one.

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

x^2 - 1 = (x -1)(x + 1)   

so the least common denominator is (x + 1)(x + 1)(x - 1) which equals x^3 + x^2 - x - 1

So now you need to change each fraction just as you would with 1/4 + 1/3 = 3/12 + 4/12 = 7/12

You need to multiply the first fraction (both the numerator and denominator each) by (x - 1)/(x -1).

((x-1)/(x-1))  ((3x + 3)/(x^2 +2x + 1)) =  (3x^2 + 6x + 3)/(x^3 + x^2 - x - 1)   which is our new first team, which has the same value as the original since (x-1)/(x-1) equals one.

Second term needs (x + 1) so:  ((x+ 1)/(x + 1)) ((x - 1)/(x^2 - 1) = (x^2 - 1)/(x^3 + x^2 - x - 1).

New addition: (3x^2 + 6x + 3)/(x^3 +x^2 -x -1) + (x^2 - 1)/(x^3 +x^2 -x -1) = (4x^2 + 6x + 2)/(x^3 +x^2 -x -1)

Now, this is our sum...however always check if it can be reduced by factoring both the numerator and denominator as much as possible.  The denominator is just (x + 1)(x + 1)(x - 1) that we found earlier. 

The numerator has a GCF of 2:     2( 2x^2 + 3x + 1)

Factor by grouping:   2 ( 2x^2 + 2x + x + 1)

                                       2 ( 2x( x + 1) + 1( x + 1))

                                      2 ( 2 x + 1 ) ( x + 1).  

                                The numerator and denominator have a common factor of (x + 1). These will cancel out and we are left with 2 (2x + 1) / (x^2 - 1 ). We can distribute the numerator or leave it as it is, it doesn't matter.

Answer 4

Daisy,

I'm not exactly clear what rational expressions you meant to add. I realize getting the question written down and communicated is a bit of a challenge with a keyboard, but it makes our time much more effective. This was the expression that I gathered from your post:

(3x + 3/x^2 + 2x + 1) + (x - 1/x^2 - 1)^2

Taking into account the order of operations, start by squaring the second term:

(x - 1/x^2 - 1)
(x - 1/x^2 - 1)
---------------
-x + 1/x^2 + 1
- 1/x + 1/x^4 + 1/x^2
+ x^2 - 1/x - x

After collecting like terms:

x^2 - 2x - 2/x + 2/x^2 + 1/x^4 + 1

Add the first term to the now squared second term, and collect like terms:

x^2 - 2x - 2/x + 2/x^2 + 1/x^4 + 1 + 3x + 3/x^2 + 2x + 1

x^2 + 3x - 2/x + 5/x^2 + 1/x^4 + 2

The common denominator among the various terms is x^4, by inspection, so one should select various representations of 1 (i.e. x^4/x^4, x^3/x^3, and so on) in order to produce the desired common denominator for each term and multiply:

(x^4/x^4)(x^2 + 3x + 2) - (x^3/x^3)2/x + (x^2/x^2)5/x^2 + 1/x^4

x^6/x^4 + 3x^5/x^4 + 2x^4/x^4 - 2x^3/x^4 + 5x^2/x^4 + 1/x^4

And finally collect like terms once again:

(x^6 + 3x^5 + 2x^4 - 2x^3 + 5x^2 + 1)/x^4

The resulting expression, as was the initial expression, is undefined for the value x = 0.