Solve the rational equation:

\[1-\frac{6}{x} = -\frac{8}{x^2}\]

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:


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.


The first step is determining the values that cancel the denominator: \[x = 0\]

The value \(x = 0\) cannot be considered an acceptable solution because settings the denominators to zero would result in an indeterminate form. So, the equation is defined in the following range: \[D = \mathbb{R} – \left\{ 0 \right\}\]

Now, we must find the least common denominator of all denominators. We have: \[\frac{x^2-6x+8}{x^2}=0\]

We multiply both sides by \(x^2\) to simplify denominators (remember to be very careful when simplifying a variable from an equation): \[\cancel{{\color{gray}{x^2}}} \cdot \frac{x^2-6x+8}{\cancel{{\color{gray}{x^2}}}}=0 \cdot \cancel{{\color{gray}{x^2}}} \]

We have obtained a quadratic equation in its standard form \(ax^2+bx+c=0\). We can solve it using the quadratic formula or by factoring the associated polynomial. Using the quadratic formula, we get: \begin{align*} x_{1,2} &= \frac{6 \pm \sqrt{(-6)^2-4(1)(8)}}{2} \\ &= \frac{6 \pm \sqrt{36-32}}{2} \\ &= \frac{6 \pm \sqrt{4}}{2} \\ \end{align*}

We have: \[\ x = \frac{6 + \sqrt{4}}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4 \] \[x = \frac{6 – \sqrt{4}}{2} = \frac{6 – 2}{2} = \frac{4}{2} = 2 \]

Both solutions are acceptable since they differ from \(x = 0 \) and do not cancel the denominator.

The solution to the equation is: \[x_1=2 \quad x_2=4\]