Question - Combining Fractions with Common Denominator
Solution:
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 $$x = -3$$.
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