Solve rational equations and inequalities

Rational equations and inequalities concern values in the domain of a quotient of polynomial expressions: denominator zeros are excluded, and transformations involving denominators must preserve restrictions and, for inequalities, account for sign. Solving them involves finding zeros and undefined points, analyzing the quotient’s sign on the resulting intervals, and representing solution sets with interval or set notation; this connects algebraic solutions to where a rational-function graph meets or lies above or below the $x$ -axis. The scope is standard polynomial rational expressions, not abstract or more advanced generalized forms.

Detailed Explanation: Solve rational equations and inequalities

To solve a rational inequality, follow these steps:

State the domain restrictions: Find values that make a denominator zero. These values are never part of the solution.
Find the zeros of the numerator: These are values where the rational expression equals $0$ .
Place all critical values on a number line. They divide the number line into intervals.
Test the sign of the expression in each interval.
Include or exclude endpoints based on the inequality symbol and domain restrictions.

Example

Solve

\frac{(x-2)(x+1)}{x-3}\ge 0.

1. Find the domain restriction

The denominator cannot equal zero:

x-3\ne 0 \quad\Rightarrow\quad x\ne 3.

Thus, $x=3$ must be excluded from the solution.

2. Find the numerator zeros

Set each numerator factor equal to zero:

x-2=0\Rightarrow x=2,

x+1=0\Rightarrow x=-1.

The critical values are $-1$ , $2$ , and $3$ .

3. Test the intervals

These values divide the number line into four intervals:

(-\infty,-1),\qquad (-1,2),\qquad (2,3),\qquad (3,\infty).

Test one value from each interval.

\begin{array}{c|c} \text{Interval} & \text{Sign of } \dfrac{(x-2)(x+1)}{x-3}\\ \hline (-\infty,-1) & -\\ (-1,2) & +\\ (2,3) & -\\ (3,\infty) & + \end{array}

Because the inequality is $\ge 0$ , we want the intervals where the expression is positive, along with the zeros of the numerator.

So we include:

$(-1,2)$ , where the expression is positive;
$(3,\infty)$ , where the expression is positive;
$x=-1$ and $x=2$ , where the expression equals $0$ .

We exclude $x=3$ because it makes the denominator zero.

Therefore, the solution is

\boxed{[-1,2]\cup(3,\infty)}.

For a rational equation, use the same restrictions and numerator zeros, but keep only values that make the expression equal to the required number. Never include a value that makes a denominator zero.

Learn by doing: Solve rational equations and inequalities

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Rational Function Inequalities - Expanded Quadratic over Binomial - Solution Set

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