High School Algebra II Unlocked (2016)

Lesson 3.2. Rational Inequalities in One Variable

REVIEW

If you multiply or divide both sides of an inequality by a negative number, you must switch the direction of the inequality sign.

If you rewrite an inequality statement with the sides reversed, you must switch the direction of the inequality sign; if a > b, then b < a.

When graphing an inequality solution set on a number line, use a solid circle to indicate that an endpoint is included in the solution set (corresponding to the signs ≥ and ≤) and an open circle to indicate that an endpoint is not included in the solution set (corresponding to the signs > and <).

A rational inequality is an inequality that contains one or more rational expressions. Be sure to follow the same rules for inequality signs that you would with numerical inequalities, when dealing with algebraic expressions.

Kristi is going for a one-hour run. She covered the first mile in 8 minutes. How many more miles must she run if she wants to average less than 11 minutes per mile for the entire run?

The key words “less than” mean that we should set up an inequality to solve this problem. Kristi wants her average to be less than 11 minutes per mile, so the inequality will end in < 11. Her average minutes per mile is her total number of minutes spent running divided by her total number of miles run. Her total number of minutes is 60, because she is going for a one-hour run and there are 60 minutes in one hour. The unknown, which we can call m, is the number of miles left in her run, but she has already run one mile. So, her total number of miles in the run is m + 1. Let’s write our inequality.

< 11

Notice that the fact
that her first mile
took 8 minutes
is extraneous
information—not
necessary for writing
or solving this
particular inequality.
Some tests provide
extraneous information
to throw you off.

Solve the inequality for m.

The solution for m is m > 4 5/11. Kristi must run more than 4 5/11 more miles during this run if she wants to average less than 11 minutes per mile for the entire run.

The rational inequality in Example 5 was simple to solve algebraically, but most rational inequalities cannot be solved solely using the processes we use for rational equations.

To solve a rational inequality, rewrite as a rational expression set across an inequality symbol from 0, find the zeros and points of discontinuity of the rational expression, and find which intervals between those points make the inequality true.

Solve the following inequality.

One way to approach this is to temporarily solve as an equation, cross-multiplying to get x² − x = 2x + 10, moving all terms to one side to get x² − 3x − 10 = 0, and solving for the zeros of the polynomial. But we would need to also remember that x ≠ −5 and x ≠ 1 from the original equation and use these values of x along with the zeros as points to divide the number line into intervals. Another approach is to subtract from both sides.

The rational expression will equal 0 if either factor in the numerator equals 0. When x + 2 = 0, x = −2, and when x − 5 = 0, x = 5. If either factor in the denominator equals 0, then the entire denominator equals 0 and the rational expression is undefined. So, x ≠1 and x ≠ −5. The two values that make the expression equal 0 and the two values that make it undefined are the four x-values that define the intervals we will look at. Let’s plot these four points on a number line. The points at −5 and 1 must be open circles, because these values of x are not included in the solution set. The points at −2 and 5 are closed circles, because these endpoints are included in the solution set by the ≥ sign (as opposed to the > sign).

Let’s test a value from each interval in the original inequality.

You can choose any
value you like from
each of the intervals,
so it is helpful to
choose values that
will provide simpler
calculations. For
example, testing
x = 10 produces an
easier comparison than
testing x = 8, which
would give us the
inequality 8/13 ≥ 2/7.

The inequality is true for x-values less than −5, greater or equal to −2 but less than 1, and greater than 5. Now we can graph the full solution set, as shown on the number line below.

We can also write the solution set as “x < −5, −2 ≤ x < 1, or x ≥ 5.”