Solve the rational equation:

\[\frac{2x+1}{6} = \frac{1}{x}\]


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 denominator to exclude them from the solutions of the equation: \[x = 0\]

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


Now, we must find the least common denominator of all denominators. We get: \[\frac{x(2x+1)}{6x} = \frac{6}{6x}\]

The equation becomes: \[\frac{x(2x+1)}{6x}-\frac{6}{6x} = 0\]


Now, we multiply both sides by \(6x\) to simplify denominators (remember to be very careful when simplifying a variable from an equation): \begin{align*} & \cancel{{\color{gray}{6x}}} \cdot \frac{x(2x+1)}{\cancel{{\color{gray}{6x}}}}-\frac{6}{\cancel{{\color{gray}{6x}}}} = 0 \cdot \cancel{{\color{gray}{6x}}} \end{align*}

  • We obtain: \begin{align*} & x(2x+1)-6 = 0 \\ & 2x^2 + x-6 = 0 \\ & x^2 + x-6 = 0 \end{align*}

We have obtained a second-degree equation in its standard form \(ax^2+bx+c=0\). We can solve it using the quadratic formula or by factoring the associated polynomial. Since the polynomial is easy to decompose, we can use the factoring method to find the solutions to the associated equation quickly.


The polynomial is decomposable at a glance into \((x-2)(x+3)\). If you are unfamiliar with factorisation of polynomials, review the theory and follow the procedure explained below.

To factor the polynomial \(2x^2 + x – 6\) we must find two numbers \(r_1,r_2\), whose sum \(S = r_1 + r_2\) equals \(b=1\), and whose product \(P=r_1 \cdot r_2\) equals \(a \cdot c = -6\). We can use this simple table to find the numbers that satisfy our constraints: \begin{array}{rrrr} & r_1 & r_2 & P & S \\ \hline & 2 & -3 & -6 & -1 \\ & -2 & 3 & -6 & 1\\ \end{array}


The numbers \(r_1, r_2\) satisfying the constraint are -2 and 3. The equation becomes: \[(x-2)(x+3) = 0\]

We have: \[x-2 = 0 \to x=2\] \[x+3 = 0 \to x= -3\]


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

The solution to the equation is: \[x_1=2 \quad\quad x_2=-3\]