Adding & Subtracting Rational Expressions

[malelia123]

Got another one for you!

Simplify

6 / (X^2-1) - 1 / (x+1)

Here is where i've gotten to so far.
Common denominator is
(x^2-1)(x+1)

leaving the new equations to be:

6x + 1x +6 - 1 / (x^2-1)(x+1)

answer in the text is:
x+5 / (x+1)(x-1)

HELP! I don't understand what I've done wrong and I can't find any examples where there is a power on the bottom that doesn't equal out (?)
ex. x^2 - 25 = (x+5)(x-5)

Thanks in advance!

 

[DrPhil]

Got another one for you!

Simplify

6 / (X^2-1) - 1 / (x+1)

Here is where i've gotten to so far.
Common denominator is
(x^2-1)(x+1) This is a common denominator, but not the LOWEST CD.

leaving the new equations to be:

6x + 1x +6 - 1 / (x^2-1)(x+1)

answer in the text is:
x+5 / (x+1)(x-1)

HELP! I don't understand what I've done wrong and I can't find any examples where there is a power on the bottom that doesn't equal out (?)
ex. x^2 - 25 = (x+5)(x-5)

Thanks in advance!

Be sure to factor all denominators to find common factors. Try again, using (x+1)(x-1) as the LCD.

 

[malelia123]

Be sure to factor all denominators to find common factors. Try again, using (x+1)(x-1) as the LCD.

Thanks, Any chance you could show me how to break that out to make it (x+1)(x-1)
I'm confused on how to get to that part.

 

[DrPhil]

Thanks, Any chance you could show me how to break that out to make it (x+1)(x-1)
I'm confused on how to get to that part.

ok look at the product of those two factors (use FOIL):

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

This can be read forwards of backwards:
"The product of the sum and difference of the same two numbers is the square of the first minus the square of the second."
OR
"The difference of two squares is equal to the product of the sum and difference of the square roots."

6 / (X^2-1) - 1 / (x+1) = [6 - (x - 1)] / [(x + 1)(x - 1)]

This does not give the answer in the text - could it be that the problem has + instead of - for the second fraction?

 

[Lexadis]

Thanks, Any chance you could show me how to break that out to make it (x+1)(x-1)
I'm confused on how to get to that part.

To find the least common denominator of \(\displaystyle (x^{2}-1)\) and \(\displaystyle (x+1)\), you have to do a small work :]
You know that since \(\displaystyle (x+1)\) is a neat factor, you have nothing to do, so you keep it as it is.
And then, \(\displaystyle (x^{2}-1)\) seems a bit untidy, so we have to clean it up in a much more neater way so as to make it easier to solve.

So, we make 1 as a power of 2 too. (any power to 1 is always one, right?) So,

\(\displaystyle (x^{2}-1)=\left (x^{2}-1^{2} \right )\)

Remember the formula \(\displaystyle a^{2}-b^{2}=(a-b)(a+b)\)?
We apply the same theory here:

\(\displaystyle (x^{2}-1)=\left (x-1 \right )\left ( x+1 \right )\)

Simple! Now the denominator would be \(\displaystyle \left (x-1 \right )\left ( x+1 \right )\) ^^