Linear and compound inequalities
Inequalities compare quantities that need not be equal. Solving them is almost like solving equations, with one crucial difference when multiplying or dividing by a negative number.
Inequality symbols
The symbols \(<\), \(>\), \(\le\), and \(\ge\) mean less than, greater than, less than or equal to, and greater than or equal to. An inequality such as \(x > 3\) has infinitely many solutions: every number greater than 3. Statements like 'at least 18' mean \(\ge 18\), and 'no more than 50' means \(\le 50\).
Graphs and interval notation
Solutions are shown on a number line with an open circle for a strict inequality and a closed circle when the endpoint is included. Interval notation uses brackets for included endpoints and parentheses for excluded ones: \(x > 3\) is \((3, \infty)\), and \(-2 \le x < 5\) is \([-2, 5)\). Infinity always gets a parenthesis.
Solving linear inequalities
Use the same steps as for equations. Solving \(3x - 4 < 11\): add 4 to get \(3x < 15\), then divide by 3 to get \(x < 5\). Adding or subtracting any number, or multiplying or dividing by a positive number, keeps the inequality's direction.
Flipping the sign
Multiplying or dividing both sides by a negative number reverses the inequality. From \(-2x \ge 8\), dividing by -2 gives \(x \le -4\). To see why, start with the true statement \(2 < 5\) and multiply by -1: \(-2 > -5\), so the direction must flip.
Compound inequalities with 'and'
An 'and' inequality requires both conditions, giving the overlap of their solutions. \(-3 < 2x + 1 \le 9\) can be solved in one go by operating on all three parts: subtract 1 to get \(-4 < 2x \le 8\), then divide by 2 to get \(-2 < x \le 4\), or \((-2, 4]\).
Compound inequalities with 'or'
An 'or' inequality requires at least one condition, giving the union of their solutions. \(x < -1\) or \(x \ge 3\) is written \((-\infty, -1) \cup [3, \infty)\). If two 'and' conditions do not overlap, such as \(x < 1\) and \(x > 4\), there is no solution.
Inequalities in applications
Inequalities model budgets, limits, and requirements. If a course grade needs an average of at least 80 over five tests and you have 72, 85, 90, and 78, then \(\frac{72 + 85 + 90 + 78 + x}{5} \ge 80\), so \(325 + x \ge 400\) and you need \(x \ge 75\) on the last test.
Worked example
A delivery van can carry at most 1,200 kg. The driver weighs 80 kg and each box weighs 35 kg. The number of boxes b must satisfy \(80 + 35b \le 1200\), so \(35b \le 1120\) and \(b \le 32\). The van can carry at most 32 boxes; since b counts boxes, it must also be a whole number.
Key takeaways
Inequalities have infinitely many solutions, shown on a number line with open or closed circles, or in interval notation with parentheses for excluded endpoints and brackets for included ones. Solve them like equations, but reverse the direction whenever you multiply or divide by a negative number. 'And' compound inequalities give the overlap of solutions and can be solved on all three parts at once; 'or' inequalities give the union. Translate phrases such as 'at least' and 'at most' carefully in applications.
Flashcard
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Check your understanding
Solve \(-2x \ge 8\).