To solve the equation [tex]\(\frac{2}{x-1} + \frac{3}{x+1} = \frac{-6}{x^2-1}\)[/tex], let's follow a step-by-step approach.
1. Rewrite the equation:
[tex]\[
\frac{2}{x-1} + \frac{3}{x+1} = \frac{-6}{x^2-1}
\][/tex]
2. Factor the denominator on the right-hand side:
Notice that [tex]\(x^2 - 1\)[/tex] can be factored as [tex]\((x-1)(x+1)\)[/tex]. So the equation becomes:
[tex]\[
\frac{2}{x-1} + \frac{3}{x+1} = \frac{-6}{(x-1)(x+1)}
\][/tex]
3. Combine the fractions on the left-hand side:
To combine the fractions on the left-hand side, find a common denominator, which is [tex]\((x-1)(x+1)\)[/tex]:
[tex]\[
\frac{2(x+1) + 3(x-1)}{(x-1)(x+1)} = \frac{-6}{(x-1)(x+1)}
\][/tex]
4. Simplify the numerator on the left-hand side:
Distribute and combine like terms in the numerator:
[tex]\[
\frac{2(x+1) + 3(x-1)}{(x-1)(x+1)} = \frac{2x + 2 + 3x - 3}{(x-1)(x+1)} = \frac{5x - 1}{(x-1)(x+1)}
\][/tex]
So the equation now is:
[tex]\[
\frac{5x - 1}{(x-1)(x+1)} = \frac{-6}{(x-1)(x+1)}
\][/tex]
5. Set the numerators equal to each other:
Since the denominators are the same, we can set the numerators equal:
[tex]\[
5x - 1 = -6
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Add 1 to both sides of the equation:
[tex]\[
5x = -5
\][/tex]
Divide by 5:
[tex]\[
x = -1
\][/tex]
7. Check for excluded values:
Before declaring [tex]\(x = -1\)[/tex] as a solution, we need to check whether this value is allowed in the original equation. The original equation has denominators [tex]\(x-1\)[/tex] and [tex]\(x+1\)[/tex], which means [tex]\(x \neq 1\)[/tex] and [tex]\(x \neq -1\)[/tex]. Our solution [tex]\(x = -1\)[/tex] is thus excluded because it makes the denominators [tex]\(x+1\)[/tex] zero.
Since [tex]\(x = -1\)[/tex] is not valid, and there are no other potential solutions from our process, we conclude that the original equation has no solution.
Therefore, the equation [tex]\(\frac{2}{x-1} + \frac{3}{x+1} = \frac{-6}{x^2-1}\)[/tex] has no solutions.
