# Solving Linear Inequalities

Here is an example of solving a rational inequality:

\(\dfrac{x^2+3x+2}{x^2-25}\)\(>0\)

**First,** factor the numerator and denominator.

\(\dfrac{(x+2)(x+1)}{(x+5)(x-5)}\)\(>0\)

**Second,** note that if the numerator is zero, then the whole fraction is zero. Set the numerator to 0 and solve.Therefore, \(x=-2\) or \(x=-1\) will make the fraction zero. If the fraction is zero, then the inequality is not true.

**Third,** any values that make the denominator 0, make the fraction undefined at that point. By setting the denominator equal to 0, we can see that the fraction is undefined at \(x=5\) and \(x=-5\).

**Fourth,** there is enough information to determine the intervals of the inequality. These intervals are:

\((-\infty,-5),\,(-5,-2),\,(-2,-1),\,(-1,5),\,(5,\infty)\)

Note that the parenthesis reflect that the grouping includes the numbers in the interval, but does not include the values that make the fraction equal to zero or undefined.

**Last**, it is time to find which intervals satisfy the inequality. There are two ways to do this.

The first way is to set up a table:

Solve each factor and determine which are positive and which are negative a which interval.

For example: \(x+5>0\)

\(x>-5\). Everywhere in the chart where x is greater than -5 will give a solution greater than 0.

Complete for the rest of the factors.

\(x+5\) | – | + | + | + | + |

\(x+2\) | – | – | + | + | + |

\(x+1\) | – | – | – | + | + |

\(x-5\) | – | – | – | – | + |

-1 | \(-\infty\) | 5 | -5 | \(\infty\) | -2 |

rational | + | – | + | – | + |

Then, take the product of the signs and place in the last row.

Since the inequality is greater than zero, we want all of the intervals that are positive. We do not include the values that make the fraction zero.

In interval notation, the solution is

\((-\infty,-5),\,(-2,-1),\,(5,\infty)\)

The second way to find intervals is to use test values on the number line.

Test Value | Solution | Sign of Interval |
---|---|---|

\(-6\) | \(+\dfrac{28}{11}\) | Positive |

-3 | \(-\dfrac{1}{8}\) | Negative |

-1.5 | \(\dfrac{25}{2275}\) | Positive |

3 | \(-\dfrac{10}{3}\) | Negative |

6 | \(\dfrac{56}{11}\) | Positive |

These answers confirm the intervals from the first method.

Here are two videos of additional worked examples. The first example has zero on one side and the second example does not have zero on either side. Once you have viewed the videos, return to this page by using the back arrow.

Rational inequalities: one side is zero (video) – Khan Academy

Rational inequalities: both sides are not zero (video) – Khan Academy