# Rational Equation A4

Solve the rational equation:

\[\frac{1}{x+2}-\frac{1}{x-1} = \frac{2}{x^2-1}\]

Rational equations feature at least one fraction in which the numerator and denominator are polynomials. Such equations are categorized as rational because they can be expressed as the ratio of two polynomials. Specifically, rational equations have the following form:

\[\frac{P(x)}{Q(x)}=0\]

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) \neq 0\).

The first step to solving rational equations is determining the values that make the denominators zero. These values are not allowable solutions because they lead to an indeterminate form.

The sequent step entails determining the least common multiple of the polynomials in all the denominators and finding the solutions for the polynomials in the numerator.

In the final stage of the process, removing the values that nullify the denominators and, subsequently, validating the acceptability of the remaining solutions to evaluate their admissibility in light of the given conditions is necessary.

## Solution

The first step is determining the values that cancel the denominators to exclude them from the solutions to the equation. We have: \begin{align*} x-2 = 0 \to x &= 2 \\ x-1 = 0 \to x &= 1 \\ x^2 -1 = 0 \to x &= \pm 1 \end{align*}

These values cannot be considered acceptable solutions because setting the denominators to zero would result in an indeterminate form. So, the equation is defined in the following range: \[D = \mathbb{R} – \left\{ -1, 1, 2 \right\}\]

Now we must find the least common denominator of all denominators. The polynomial \(x^2-1\) is a notable product we can factorise as \((x+1)(x-1)\). We get: \[\frac{1}{x+2}-\frac{1}{x-1} = \frac{2}{(x+1)(x-1)}\]

The equation becomes:\[ \frac{x^2-1 -(x+2)(x+1)-2(x+2)}{(x+2)(x+1)(x-1)} = 0\]

Simplifying the denominator (remember to be very careful when simplifying a variable from an equation) we obtain: \begin{align*} &x^2-1 -(x^2+x+2x+2)-2x+4 = 0\\[0.6em] &-5x-7 = 0 \\[0.6em] &x =-\frac{7}{5} \end{align*}

The solution is acceptable since it differs from \(x = 2, x= \pm 1 \) and does not cancel the denominator. The solution to the equation is: \[x=-\frac{7}{5}\]