Combining rational expressions with unlike denominators

Key takeaway: This problem asks you to combine two fractions with different denominators, then simplify the result carefully using a common denominator.

Find a common denominator

We have

$\frac{2}{x+1} - \frac{1}{x-1}$.

The least common denominator is

$(x+1)(x-1) = x^2 - 1$.

Rewrite each fraction using that denominator:

$\frac{2}{x+1} = \frac{2(x-1)}{(x+1)(x-1)}$

$\frac{1}{x-1} = \frac{x+1}{(x-1)(x+1)}$

Subtract the numerators

Now combine them:

$\frac{2(x-1) - (x+1)}{x^2 - 1}$

Distribute the minus sign carefully:

$\frac{2x - 2 - x - 1}{x^2 - 1}$

Simplify the numerator:

$\frac{x - 3}{x^2 - 1}$

So the correct answer is

$\boxed{\frac{x - 3}{x^2 - 1}}$, which is choice (c).

Why the sign step matters

The most important algebra step is subtracting the entire second numerator, not just the first term. The minus sign must affect both $x$ and $1$ in $(x+1)$.

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Pitfalls the pros know 👇 Students often make the numerator $2(x-1)-x+1$ instead of $2(x-1)-(x+1)$. That sign mistake changes the entire answer. Another common issue is using the denominator $x^2+1$ because they multiply $(x+1)(x-1)$ incorrectly. Remember that difference of squares gives $x^2-1$, not $x^2+1$. Also note the restrictions: $x$ cannot be $1$ or $-1$.

What if the problem changes? If the problem were $\frac{2}{x+1} + \frac{1}{x-1}$, the same common denominator $x^2-1$ would be used, but the numerators would add instead of subtract. The result would be $\frac{2(x-1)+(x+1)}{x^2-1} = \frac{3x-1}{x^2-1}$. If the denominators were $x+2$ and $x-2$, the common denominator would become $x^2-4$.

Tags: common denominator, difference of squares, rational expression addition