How do you solve the rational equation:
(n/n-2)+(n/n+2)=(n/n^2-4)
I know that the LCD of the equation is (n+2)(n-2) because that is what you get when you factor out the n^2-4 in the last part of the question. I thought I did it right until I reached the end where I don't know how to solve it.
So why not multiply each of the three terms by that LCD that you found to get
n(n+2) + n(n-2) = n
expand and solve as a quadratic.
I factors very nicely and has 2 simple answers.
That's just it. I don't know how to factor it out because i keep getting 2n^2-n=0
2n^2 - n = 0 is correct.
Now take out a common factor
n(2n-1) = 0
n=0 or n = 1/2
To solve the rational equation (n/(n-2)) + (n/(n+2)) = (n/(n^2-4), you correctly identified the least common denominator (LCD) as (n+2)(n-2), which is obtained by factoring n^2-4.
Now, let's solve the equation step by step:
1. Multiply each term in the equation by the LCD (n+2)(n-2) to eliminate the denominators:
(n/(n-2)) * (n+2)(n-2) + (n/(n+2)) * (n+2)(n-2) = (n/(n^2-4)) * (n+2)(n-2).
After multiplying, the equation becomes:
n(n+2) + n(n-2) = n.
2. Simplify each term:
n^2 + 2n + n^2 - 2n = n.
3. Combine like terms:
2n^2 = n.
4. Move all terms to one side to set the equation to zero:
2n^2 - n = 0.
5. Factor the equation:
n(2n - 1) = 0.
6. Set each factor equal to zero and solve for n:
n = 0
2n - 1 = 0
n = 1/2.
Therefore, the solutions to the rational equation are n = 0 and n = 1/2.