Exercise 32 Page 695

We are asked to write a rational equation that has $5$ as a solution. Let's start by recalling a few important definitions.

• Rational Equation: an equation that contains at least one rational expression.
• Rational Expression: a fraction whose numerator and denominator are polynomials.
• Extraneous Solution: a solution that is derived from the original equation, but it is not actually a solution.

With these definitions in mind, we can write the equation! First, let's arbitrarily choose the variable that we will use to write the equation. We will let $x$ be our variable. We will create our equation by starting from the solution that we want it to have. In order to obtain a rational equation, we have to use the Properties of Equality to perform the same operations on both sides of the equation. Let's start by adding $x$ and subtracting $6$ on both sides. Remember that we must have at least one rational expression in our equation. Therefore, we will divide both sides of the equation by a polynomial. Let $x^2-1$ be our polynomial. Note that we have to divide the equation by a polynomial that is not equal to $0$ when $x=5.$ Otherwise, $x=5$ will be an extraneous solution. We can further simplify the right-hand side of the equation by factoring the denominator and canceling any common factors. We have obtained a rational equation that has $5$ as a solution. To be sure, let's check that $x=5$ is actually a solution of the equation. This can be done by substituting $5$ for $x$ in the equation and simplifying. Since the obtained statement is true, $x=5$ is a solution of our rational equation. Please note that there are infinitely many rational equations that have $5$ as a solution, so our answer is only an example. Here are some other examples.

Examples
$\frac{x-5}{3x-4}=x^2-4x-5$
$\frac{2x-10}{x^2+1}=\frac{x^2-25}{x^3-x^2+3}$
$\frac{x+1}{2x^2-6}=\frac{x^2-22}{x^2-2x+7}$