Combining Fractions with Common Denominator
To solve the equation given in the image, you need to combine the fractions on the left hand side of the equation. The first step is to find a common denominator for both fractions.
The two denominators are \(x - 2\) and \(x^2 - 4\). Notice that \(x^2 - 4\) is a difference of two squares and can be factored as \((x + 2)(x - 2)\).
Thus, the least common denominator (LCD) for both fractions is \(x^2 - 4\) or \((x + 2)(x - 2)\).
Now write both fractions with the common denominator:
\[\frac{x + 1}{x - 2} = \frac{(x + 1)(x + 2)}{x^2 - 4}\]
The second fraction is already with the denominator \(x^2 - 4\), so leave it as is:
\[\frac{x}{x^2 - 4}\]
Now that you have a common denominator, you can combine the numerators:
\[\frac{(x + 1)(x + 2) + x}{x^2 - 4}\]
Expand the numerator:
\[\frac{x^2 + 2x + x + 2 + x}{x^2 - 4}\]
Combine like terms:
\[\frac{x^2 + 4x + 2}{x^2 - 4}\]
Set this equal to the right-hand side of the original equation:
\[\frac{x^2 + 4x + 2}{x^2 - 4} = \frac{-1}{x^2 - 4}\]
Since the denominators are the same, we can equate the numerators:
\[x^2 + 4x + 2 = -1\]
Bring the -1 to the left side of the equation:
\[x^2 + 4x + 2 + 1 = 0\]
Combine like terms:
\[x^2 + 4x + 3 = 0\]
Now, factor the quadratic equation:
\[(x + 3)(x + 1) = 0\]
Set each factor equal to zero and solve for x:
\[x + 3 = 0 \quad \text{or} \quad x + 1 = 0\]
\[x = -3 \quad \text{or} \quad x = -1\]
So the solutions for x are -3 and -1.
However, we must check for possible extraneous solutions because we're working with rational equations. Since \(x = -1\) makes the original denominators \((x - 2)\) and \((x^2 - 4)\) equal to zero, \(x = -1\) is an extraneous solution and must be discarded.
Therefore, the solution to the equation is \(x = -3\).
