Linear Inequalities in One Variable

Life is full of limits

Your pocket money is a budget. The road has a speed limit. An exam has a pass mark. In real life, conditions are far more often "at most" or "at least" than "exactly equal". That is what inequalities describe. The good news is that solving a linear inequality is almost the same as solving an equation. There is just one rule that is different, and we will keep a close eye on it. We will also learn to show the answers on a number line.

The rules

An inequality keeps its direction when you add or subtract the same number on both sides, and when you multiply or divide both sides by a positive number. Here is the one that trips people up: multiplying or dividing by a negative number reverses the sign. For example, $-2x > 6$ gives $x < -3$ . Check it with a number if you doubt it; the rule is not arbitrary.

Solving

Example 1. Solve $5x - 3 < 3x + 9$ . Collecting terms, $2x < 12$ , so $x < 6$ . The answer also depends on which numbers are allowed. For $x \in \mathbf{R}$ the solution is $(-\infty, 6)$ ; for $x \in \mathbf{N}$ it is $\{1, 2, 3, 4, 5\}$ .

Example 2. Solve $\displaystyle \frac{x - 2}{3} \ge \frac{2x + 1}{5} - 1$ . Clear the fractions by multiplying by $15$ : $5x - 10 \ge 6x + 3 - 15$ . Then $-x \ge -2$ , and flipping the sign as we multiply through by minus one, $x \le 2$ . The solution is $(-\infty, 2]$ .

Example 3. A double inequality is handled on all three parts at once. Take $\displaystyle -4 \le \frac{3 - 2x}{5} < 3$ . Multiply by $5$ : $-20 \le 3 - 2x < 15$ . Subtract $3$ : $-23 \le -2x < 12$ . Now divide by $-2$ , remembering to reverse both signs: $\displaystyle \tfrac{23}{2} \ge x > -6$ . The solution is $\displaystyle (-6, \tfrac{23}{2}]$ .

Number line from -8 to 13 with the solution of Example 3 shaded from -6 to 23/2: an open circle at -6, which is excluded, and a filled dot at 23/2, which is included. — Example 3: the solution set from -6 (excluded) to 23/2 (included)

Example 4. Sometimes the variable disappears altogether. Then look at what is left. $3(x - 1) > 3x + 2$ reduces to $-3 > 2$ , which is false, so there is no solution. $2(x + 1) \le 2x + 5$ reduces to $2 \le 5$ , which is always true, so every real $x$ works.

On the number line

A picture makes the answer clear at a glance. Put an open circle at an endpoint that is left out ( $<$ or $>$ ), a filled circle at one that is included ( $\le$ or $\ge$ ), and shade the solution. When there are two inequalities in one variable to satisfy together, solve each one and keep only the part where the two shadings overlap.

Example 5. From $2x - 1 > 3$ and $5 - x \ge 1$ we get $x > 2$ and $x \le 4$ . Together, that is $(2, 4]$ .

Three number lines from 0 to 7: the first shades x greater than 2 with an open circle at 2, the second shades x at most 4 with a filled dot at 4, the third shows their overlap (2, 4]. — Example 5: intersecting x greater than 2 with x at most 4 gives (2, 4]

Word problems

Example 6. A student has marks of $68$ and $74$ in two tests, and needs an average of at least $75$ over three. The three marks must total at least three times that, so we need $68 + 74 + x \ge 225$ , which gives $x \ge 83$ .

Example 7. Two consecutive odd natural numbers are each greater than $10$ , and their sum is less than $40$ . If the smaller one is the unknown, then $x > 10$ and $x + (x + 2) < 40$ give $10 < x < 19$ with $x$ odd, so the pairs are $(11, 13), (13, 15), (15, 17), (17, 19)$ .

Example 8. A solution in a laboratory must be kept between $20^\circ$ C and $30^\circ$ C. Using $\displaystyle F = \tfrac{9}{5}C + 32$ on both ends, the range is $68 < F < 86$ degrees Fahrenheit.

Practice questions

Solve for $x \in \mathbf{R}$ : $4x + 7 > 2x - 5$ ; $3 - 5x \le 18$ ; $\displaystyle \tfrac{x}{4} < \tfrac{5x - 2}{3} - \tfrac{7x - 3}{5}$ .
Solve for $x \in \mathbf{Z}$ : $-3 < 2x + 1 \le 9$ .
Solve and draw: $\displaystyle -5 \le \frac{2 - 3x}{4} \le 9$ .
Solve the system: $3x - 7 < 5 + x$ and $11 - 5x \le 1$ .
Solve: $\displaystyle \frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$ .
A student scored $62$ , $71$ and $79$ . What must the fourth score be for an average of at least $70$ ?
Find all pairs of consecutive multiples of $4$ , both greater than $10$ , whose sum is less than $60$ .
In a triangle, one side is twice the shortest and the third side is $4$ cm longer than the shortest. The perimeter is at most $52$ cm. Find the greatest possible length of the shortest side.
How many litres of water must be added to $600$ L of a $40\%$ salt solution so that the result is between $20\%$ and $30\%$ salt?
Explain why $\lvert x \rvert < 3$ means $-3 < x < 3$ , and solve $\lvert 2x - 1 \rvert \le 5$ .

Answers

Show answers

$x > -6$ ; $x \ge -3$ ; multiply by $60$ : $15x < 100x - 40 - 84x + 36$ , i.e. $-x < -4$ , $x > 4$ .
$-2 < x \le 4$ : $\{-1, 0, 1, 2, 3, 4\}$ .
$-20 \le 2 - 3x \le 36$ : $\displaystyle -\tfrac{34}{3} \le x \le \tfrac{22}{3}$ .
$x < 6$ and $x \ge 2$ : $[2, 6)$ .
Multiply by $60$ : $40x - 20 \ge 45x - 30 - 24 + 12x$ ; $-17x \ge -34$ ; $x \le 2$ .
$62 + 71 + 79 + x \ge 280$ : $x \ge 68$ .
$(12, 16), (16, 20), (20, 24), (24, 28)$ . The smaller one must satisfy $x > 10$ and $2x + 4 < 60$ .
$x + 2x + (x + 4) \le 52$ : $x \le 12$ cm.
Salt $240$ L; $\displaystyle 0.2 < \tfrac{240}{600 + w} < 0.3$ : $200 < w < 600$ L.
$\lvert x \rvert$ is the distance from $0$ , so it is less than three exactly on that stretch. Then $-5 \le 2x - 1 \le 5$ gives $-2 \le x \le 3$ .

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Linear Inequalities in One Variable

Life is full of lim­its#

The rules#

Solv­ing#

On the num­ber line#

Word prob­lems#

Prac­tice ques­tions#

Answers#

Test yourself