Rational Inequalities

Rational inequalities involve a ratio of two linear functions of the same variable in the form $\frac{ax+b}{cx+d}$ .

Example

infoNote

Solve for $x$ in the following inequality :

\frac{x-8}{x+2} \ge-1

Following algebraic principles, it would be intuitive to multiply by $(x+2)$ to eliminate the fraction. However we don't know if the expression $(x+2)$ is positive or negative since it depends on what $x$ value we sub in. This means that we don't know whether or not to switch the inequality sign.

A simple way-around is to multiply both sides by $(x+2)^2$ because any expression squared is guaranteed to be positive, which means we don't have to switch the inequality sign :

(x+2)^2\left( \frac{x-8}{x+2} \right) \ge-1 (x+2)^2

We can rewrite the inequality as :

\frac{(x-8)(x+2)(x+2)}{x+2} \ge-1 (x+2)^2

You should notice a common factor of $(x+2)$ which we can cancel out :

\begin{align*} \frac{(x-8)\cancel{(x+2)}(x+2)}{\cancel{(x+2)}} &\ge-1 (x+2)^2\\\\ (x-8)(x+2) &\ge-1 (x+2)^2 \end{align*}

Expand and simplify :

\begin{align*} x^2-8x+2x-16 &\ge-x^2-4x-4 \\\\ 2x^2-2x-12 &\ge 0 \end{align*}

Using the quadratic formula derives the roots : $x=3$ and $x=-2$ . Evaluate the region test just as before :

So, the solution set is :

-2\ge x\ge3.

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