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2.7 Solving Rational Inequalities

The five unequal signs

In addition to the equal sign, there are five unequal signs which stand for the different ways two quantities may be unequal. The five signs are

less than \(\quad\) \( < \)
less than or equal to \(\quad\) \(\le\)
greater than \(\quad\) \( > \)
greater than or equal to \(\quad\) \(\ge\)
not equal to \(\quad\) \(\ne\)

An inequality is a statement that two quantities are unequal in one of the five ways listed above. A solution of an inequality is a description of the set of all values of the variable for which the inequality is true.

A rational inequality is in inequality in which the two expressions are rational expressions. That is, polynomials or ratios of polynomials.

Example 2.7.a

Solve the inequality \(2 x + 3 > 9\).

As in an equation, we may subtract three from both sides is the inequality to get \(2 x > 6\)

Then we may divide by 2 to get \(x > 3\).

The solution is the set of all numbers larger than \(3\), or the interval \((3, \infty )\).

Multiplying an inequality by a negative number

Although the steps in solving the inequality in Example 2.7.a are the same steps one would take in solving \(2 x + 3 = 9\), there is one important difference when one is solving inequalities:

Multiplication or division of both sides of an inequality by a negative number reverses the direction of the inequality

To see why this must be so, consider the next example.

Example 2.7.b

Solve \(-x\ge 5\).

First, we are going to solve it without multiplying.

Add \(x\) to both sides: \(\quad\) \(0\ge x+5\)
Subtract \(5\) from both sides: \(\quad\) \(-5\ge x\)
Re-orient\(^*\) the inequality: \(\quad\) \(x\le -5\)

So the solution set is the unbounded interval \((-\infty,-5)\).

\(^*\) To re-orient an inequality, one re-writes the inequality in the equivalent but opposite sense. The inequalities \(A > B\) and \(B < A\) are equivalent but of opposite sense.

Notice that the final result \(x\ge-5\) could have been obtained from the original inequality \(-x\le5\) in one step instead of three by the simple process of multiplying both sides by \(-1\) and reversing the direction of the inequality.

Example 2.7.c

Solve the inequality: \(1 - x < 5\)

Subtract 1 from each side to get: \(- x < 4\)

Then multiply both sides by \(- 1\), remembering to reverse the direction of the inequality: \(x > - 4\)

So the solution is \(( - 4, \infty) \).

Exercise 2.7.1

Solve the linear inequality: \(3 x - 7 > 5\).

See Solution

Solving inequalities by factoring

We are going to look at a general method of solving an inequality in which one side of the inequality is zero.

Requiring that one side of the inequality equal zero is not a difficult requirement, since we can always achieve that state in one step by subtracting the expression on one side of the inequality from both sides of the inequality.

Let us reconsider the first inequality we looked at above: \(2 x + 3 > 9\)

Subtract 9 from each side to get: \(2 x - 6 > 0\)

Now factor the left side: \(2 ( x - 3 ) > 0\)

Now sketch the graph of \(y = 2 ( x - 3 )\). It crosses the \(x\)-axis at \(3\) since \(3\) is a transitive \(x\)-intercept. The rightmost part of the graph is above the \(x\)-axis since the leading coefficient, \(2\), is positive. So the graph lies above the \(x\)-axis to the right of \(3\) and the graph lies below the \(x\)-axis to the left of \(3\). Thus, we can see that \(y > 0\) when \(x > 3\).

Thus the solution of the inequality is \(( 3, \infty) \).

So the method we are going to learn for solving rational inequalities will involve getting zero on one side of the inequality and a rational expression on the other side. Then we graph the rational expression and see which portions of the graph satisfy the requirements of the inequality. The domain of those portions is the solution to the inequality.

Exercise 2.7.2

Sketch the graph of \(y = 2 - x\). Use the graph to solve the inequality \(2 - x < 0\).

See Solution

Breaking down the steps

It is not actually necessary to sketch the entire graph. All we need to know is the following:

  1. What are the critical numbers of the rational expression? That is, what values of the variable cause the expression to equal \(0\) or to be undefined? That is, what are the zeros of the numerator and what are the zeros of the denominator?
  2. Which of the critical numbers are transitive and which are intransitive? Remember that the critical number \(c\) is transitive if \( x - c \) is a factor of odd multiplicity and intransitive if \( x - c \) is a factor of even multiplicity. This is true whether \( x - c \) is a factor of the numerator or a factor of the denominator. The expression changes signs at transitive critical numbers, but not at intransitive critical numbers.
  3. Is the expression positive or is it negative to the right of the largest critical number? To find out, take the ratio of the leading coefficient of the numerator to the leading coefficient of the denominator. If the ratio is positive, then the expression is positive to the right of the largest critical number. If the ratio is negative, then the expression is negative to the right of the largest critical number.

Example 2.7.d

Solve \(\dfrac{2 x ( 1 - x )}{( x + 2 )^2} \ge 0\).

The critical numbers are \(- 2, 0\) and \(1\). The critical number \(- 2\) is intransitive and \(0\) and \(1\) are transitive. The leading coefficient of the numerator is \(- 2\) and the leading coefficient of the denominator is \(1\). So the ratio of the leading coefficients is a negative number. Thus the expression is negative to the right of \(1\), zero at \(1\), positive between \(0\) and \(1\), zero at \(0\), negative between \(- 2\) and \(0\), undefined at \(-2\) and negative to the left of \(- 2\). So the only place the expression is greater than or equal to zero is on the interval \([ 0, 1 ]\). All this can be determined without actually drawing the graph.

Exercise 2.7.3

Solve the inequality \(\dfrac{( x^2 + 1 )( 2 x - 1 )}{( x + 1 )^3( x - 1 )^2 } < 0\).

See Solution

Do not divide by expressions containing the variable.

When solving rational inequalities, be sure NOT to divide both sides by any expression containing the variable. This can result in an inequality which is not equivalent to the original, that is, an inequality which does not have the same solution as the original.

For example, solve \(\dfrac{x}{ x + 1 } < \dfrac{x }{ x + 2}\)

Tempting though it is, we do not divide both sides by \(x\). To do so would necessitate consideration of the case where \(x\) is negative apart from the case where \(x\) is positive. This would needlessly complicate the solution. Instead, we subtract one of the two expressions from both sides, find a common denominator and factor the expression.

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

\(\dfrac{x}{ ( x + 1 ) ( x + 2 )} < 0\)

The critical numbers are \(- 2, - 1\), and \(0\), all transitive. The expression is zero at \(0\) and undefined at \(- 1\) and \(- 2\). The expression is positive to the right of \(0\), negative between \(- 1\) and \(0\), positive between \(- 2\) and \(- 1\) and negative to the left of \(- 2\). Thus the expression is less than or equal to zero on the interval \(( - \infty, -2 )\) and \(( - 1, 0 ]\). So the solution is the set union of those two intervals, or \(( - \infty, - 2 ) \cup ( - 1, 0 ]\) in interval notation.

Exercise 2.7.4

Solve the inequality \(x ( x - 1 ) > \dfrac{ x - 1}{ x}\).

See Solution