Directions:Solve the rational expression.
Problem:
1/(x+3)+1/(x-3)=1/(x^2-9)
get (x+3), (x-3), and (x^2-9) to the same denominator...
To solve the rational expression, we need to get all the terms to have the same denominator. In this case, the denominators are (x+3), (x-3), and (x^2-9).
To bring them to the same denominator, we need to factor the expressions in the denominators.
First, let's factor x^2-9. This is a difference of squares, so we can rewrite it as (x+3)(x-3).
Now, we have the following denominators: (x+3), (x-3), and (x+3)(x-3).
To bring 1/(x+3) and 1/(x-3) to the same denominator as 1/(x^2-9), we need to multiply the numerator and denominator of each fraction by the missing factor(s) in the denominator.
For 1/(x+3), we need to multiply the numerator and denominator by (x-3).
So, 1/(x+3) = [(1)(x-3)]/[(x+3)(x-3)] = (x-3)/(x^2-9).
For 1/(x-3), we need to multiply the numerator and denominator by (x+3).
So, 1/(x-3) = [(1)(x+3)]/[(x-3)(x+3)] = (x+3)/(x^2-9).
Now our equation becomes:
(x-3)/(x^2-9) + (x+3)/(x^2-9) = 1/(x^2-9).
Combine the fractions on the left side:
(x-3 + x+3)/(x^2-9) = 1/(x^2-9).
Simplify the numerator on the left side:
(2x)/(x^2-9) = 1/(x^2-9).
Now that the denominators are the same, we can set the numerators equal to each other:
2x = 1.
Finally, solve for x:
2x = 1
x = 1/2.
So, the solution to the rational expression is x = 1/2.