Equations and Inequalities

Lessons

Senior 4 · Basic Mathematics

Equations and Inequalities

5 lessons

Lesson 3 of 5

Inequalities involving products and quotients

25 min

When is the average cost low enough?

A small bakery pays 20 000 Frw a day for rent and fuel, and each loaf costs 500 Frw in flour and labour. If it bakes $x$ loaves, the average cost of one loaf is

\frac{20\,000 + 500x}{x} \ \text{Frw}

(illustrative figures). The owner wants this to be at most 700 Frw, so that a price of 700 Frw covers costs. That is an inequality with the unknown in a quotient, and it cannot be solved by the rules of the last lesson alone: you may not multiply both sides by $x$ until you know its sign. This lesson gives a method that handles any product or quotient of linear factors: the sign table.

Important

Signs of products and quotients. A product or a quotient of two numbers is

positive if the two numbers have the same sign;
negative if they have opposite signs;
zero if the first factor (or the numerator) is zero. A quotient with a zero denominator is undefined.

Sign of a linear factor: $a x + b$ ( $a \ne 0$ ) is zero at $x = -\frac{b}{a}$ , has the same sign as $a$ to the right of that point, and the opposite sign to the left. For example, $3 - x$ is zero at 3, negative for $x \gt 3$ (because $a = - 1$ ) and positive for $x \lt 3$ .

Definition

Sign table: a table that records, on each interval between the zeros of the factors, the sign of every factor and of the whole product or quotient.

Method: (1) bring everything to one side so the other side is 0; (2) factorise; (3) find the zero of each factor (the critical values) and put them in order on the top row; (4) fill in the sign of each factor; (5) multiply the signs down each column; (6) read off the intervals that satisfy the inequality, deciding for each critical value whether it is included.

Worked example 1: a product

Solve $(x - 2)(x + 3) \gt 0$ .

Solution

The critical values are $- 3$ and $2$ .

$x$		$- 3$		$2$
$x + 3$	−	0	+		+
$x - 2$	−		−	0	+
$(x - 2) (x + 3)$	+	0	−	0	+

We need " $\gt 0$ ", strictly, so the zeros are excluded:

S = \,]-\infty, -3[\, \cup \,]2, +\infty[ .

Worked example 2: a quadratic written as a product

Solve (a) $(2x - 1)(3 - x) \ge 0$ (b) $x^2 - x - 6 \lt 0$ .

Solution

Zeros at $\frac{1}{2}$ and 3. $2 x - 1$ is negative then positive; $3 - x$ is positive then negative. The product is $-$ on $]-\infty, \frac{1}{2}[$ , $+$ on $]\frac{1}{2}, 3[$ , $-$ on $]3, +\infty[$ . With " $\geq \ge$ " the zeros are included: $S = \left[\frac{1}{2}, 3\right]$ .
Factorise: $(x - 3)(x + 2) \lt 0$ . The product is negative between the zeros: $S = \,]-2, 3[$ .

Worked example 3: a quotient

Solve $\frac{x + 1}{x - 4} \le 0$ .

Solution

Critical values: $- 1$ (numerator zero) and $4$ (denominator zero, so the quotient is undefined there, shown by ||).

$x$		$- 1$		$4$
$x + 1$	−	0	+		+
$x - 4$	−		−	0	+
$\frac{x + 1}{x - 4}$	+	0	−	\|\|	+

The quotient is $\le 0$ on $[- 1, 4 [$ : $- 1$ is included (the quotient equals 0 there), but 4 is excluded, whatever the inequality sign.

Worked example 4: the bakery

For which numbers of loaves $x \gt 0$ is $\frac{20\,000 + 500x}{x} \le 700$ ?

Solution

Bring everything to one side over a common denominator:

\frac{20\,000 + 500x}{x} - 700 \le 0 \iff \frac{20\,000 + 500x - 700x}{x} \le 0 \iff \frac{20\,000 - 200x}{x} \le 0 .

Critical values: 100 (numerator) and 0 (denominator). For $x \gt 0$ the denominator is positive, so we need $20\,000 - 200x \le 0$ , that is $x \ge 100$ . The bakery must bake at least 100 loaves a day for the average cost to be 700 Frw or less. (At 100 loaves: $\frac{70\,000}{100} = 700$ . ✓)

Warning

Never multiply an inequality by an expression whose sign you do not know. To solve $\frac{3x - 6}{x + 1} \ge 2$ , do not multiply by $x + 1$ : it is negative for $x \lt -1$ , which would reverse the sign there. Instead subtract 2: $\frac{3x - 6 - 2(x + 1)}{x + 1} = \frac{x - 8}{x + 1} \ge 0$ , and use a sign table: $S = \,]-\infty, -1[\, \cup [8, +\infty[$ .

Tip

With two linear factors whose $x$ -coefficients are both positive, the signs from left to right are always $+, -, +$ : the expression is positive outside the two zeros and negative between them. If one coefficient is negative, the pattern flips to $-, +, -$ . A sign table still settles every doubtful case.

The Organic Chemistry Tutor: Rational Inequalities. Watch how the critical values split the number line and why the zero of the denominator is never included.

✦

Practice

Try each one before opening the answers. They get harder as you go.

Solve $(x - 1)(x - 5) \le 0$ .
Solve $(x + 4)(2 - x) \lt 0$ .
Solve $x^2 - 9 \ge 0$ .
Solve $\frac{x - 3}{x + 2} \gt 0$ .
Solve $\frac{2x + 6}{5 - x} \ge 0$ .
Solve $\frac{x + 5}{x - 1} \le 2$ .
A printer charges 15 000 Frw to set up a job plus 200 Frw per copy. For how many copies is the average cost per copy below 350 Frw?

Show answersHide answers

Negative between the zeros: $[1, 5]$ .
Signs $-, +, -$ : $]-\infty, -4[\, \cup \,]2, +\infty[$ .
$(x - 3)(x + 3) \ge 0$ : $]-\infty, -3] \cup [3, +\infty[$ .
$]-\infty, -2[\, \cup \,]3, +\infty[$ .
Zero of numerator $- 3$ , of denominator 5: $[- 3, 5 [$ .
$\frac{x + 5 - 2(x - 1)}{x - 1} = \frac{7 - x}{x - 1} \le 0$ : $]-\infty, 1[\, \cup [7, +\infty[$ .
$\frac{15\,000 + 200n}{n} \lt 350 \iff \frac{15\,000 - 150n}{n} \lt 0$ ; with $n \gt 0$ : $n \gt 100$ , so 101 copies or more.

Quick check

What is the solution set of (x − 1)(x + 4) < 0?

Key points

Bring everything to one side, factorise, and use a sign table.
A product or quotient is positive when its factors have the same sign, negative when they differ.
Zeros of the numerator are included for $\leq \le$ or $\geq \ge$ ; zeros of a denominator are never included.
Never multiply by an expression of unknown sign.

Sources and credits

Written by Ganzaa: the explanations, the sign tables and diagrams, the worked examples, the practice items and the questions. The bakery and printer figures are illustrative.
Video: Rational Inequalities, The Organic Chemistry Tutor, embedded from YouTube.

1When is the average cost low enough?

When is the average cost low enough?