# How To Solve Algebraic Inequalities

Polynomial and rational inequalities are all about finding boundary numbers. In this document, let me show you how to do that with some examples.

**Solving Algebraic Inequalities**

The first thing you do is find the boundary numbers. These are x-values. They are obtained by working with the expression you are given. There will be some amount of arranging and factoring of the equation. From there, you can see the x-values which give you the boundaries you are looking for.

Write the equation to look like a normal equation.

Solve the equation, the solutions are the boundary numbers.

Use these boundary numbers to separate the number line into sections. At either end of the boundaries, they will be positive or negative. Whichever they are, the boundaries next to it will be the opposite sign. The boundaries will alternate signs until you get to the last boundary.

To solve the inequality, pick a number at one end of your boundaries and input it into your equation.

**Example 1**

Solving a polynomial inequality

Solve \(X^3 \leq 2X^2 + 3X \)

Begin by rewriting the inequality. It becomes:

\[ x^3 -2x^2 -3x \leq 0 \]

Replace the \(\leq\) symbol with an equal sign and solve the resulting equation.

\[ x^3 -2x^2 -3x = 0 \]

\[ x(x^2-2x-3) =0\]

\[x(x+1)(x-3) =0\]

\[x=0, x=1, x=3\]

We have three boundary numbers. When you put these points on a number line, they give you four intervals.

\[(-\infty,-1) (-1,0) (0,3) (3,\infty)\]

Take a test value from one of the intervals and note whether it is positive or negative. The boundary values alternate between positive and negative. The original question asks where it is positive or equal to zero. So we want those boundaries.

The solution is:

\[[-1,0] \cup [3,8] \]

**Rational Inequalities**

To solve rational inequalities, we can use the same basic techniques that we used to solve polynomial inequalities. There is one important exception. Boundary numbers also occur at x-values where the denominator of any rational expression in the equation equals zero.

**Solving Rational Inequalities**

When solving a rational inequality, it is essential not to multiply or divide each side of the inequality by the least common denominator if it contains a variable. This technique will lead to a wrong answer.

**Example 2**

Solve \[\frac{2-x}{2x} > 0 \]

Since the inequality is already written in the correct form, we can just solve.

The numerator is:

\[ 2 – x = 0 \]

\[ x = 2 \]

The denominator is:

\[ 2x = 0 \]

\[ x = 0 \]

The boundary numbers are 2 and 0, which separate our number line into three intervals:

\[ (-\infty,0) (0,2) (2, \infty) \]

Take a test value from one end of the intervals. Whatever sign it is, the intervals next to it alternate.

Let us use \(-2\)

\[ \frac{2-(-2)}{2(-2)} > 0 \]

\[ \frac{4}{-4} \]

\[= -1 \]

Since this end is negative, the others just alternate signs.

**Example 3**

Solve \(\frac{1}{x} \leq \frac{2}{x+1} \)

Begin by writing the inequality in the correct form.

\[ \frac{1}{x} - \frac{2}{x+1} \leq 0 \]

\[ \frac{(x+1) – 2x}{x(x+1)} \leq 0 \]

\[ \frac{-x+1}{x(x+1)} \leq 0 \]

\[ \frac{1-x}{(x)(x+1)} \leq 0\]

Find the zeros of the numerator and denominator.

Numerator:

\[ 1 – x = 0 \]

\[ -x = -1 \]

\[ x = 1 \]

Denominator:

\[ x = 0 \]

\[ x + 1 = 0 \]

\[ x = -1 \]

Our zeros are -1, 0, 1. So our boundaries are:

\[ (-\infty,-1) ( -1, 0) ( 0, 1) (1, \infty) \]

Now that we have boundaries, pick a test number at one end of the intervals. We will pick -2.

\[ \frac{1-x}{x(x+1)} \]

\[ \frac{1-(-2)}{(-2)(-2+1)} \]

\[ \frac{3}{(-2)(-1)} \]

\[ =\frac{3}{2} \]

This is positive. So, our boundaries alternate signs.

**Conclusion**

In this paper, we have talked about the differences between polynomial and rational inequalities. Then I showed examples of solving polynomial inequalities. After that, we talked about the steps to solve rational inequalities. Finally, we did some examples of ration inequalities.