Solving Linear Inequalities

How to solve linear inequalities such as 2x − 3 ≥ 5, 7 − 2x ≥ 1, 5x − 3 > 2x + 9 and 3(x − 2) ≤ 9, check the answer, and show it on a number line.

Before you start

Adding or subtracting the same number on both sides of an inequality does not change the sign. Neither does multiplying or dividing both sides by the same positive number.

Multiplying or dividing both sides by the same negative number reverses the sign, and the reversed sign is written in the same line: $-3x \le 15$ gives $x \ge -5$ .

From Secondary 1 you can solve linear equations with brackets and with fractions, such as $3(x - 2) = 9$ and $\frac{2x - 1}{3} = 5$ .

The Method

A linear inequality in $x$ is an inequality in which $x$ appears only to the power 1, such as $2x - 3 \ge 5$ . It is solved with the same steps as a linear equation. Only one step can be different: dividing or multiplying by a negative number.

Solving a linear inequality

Step 1: Expand any brackets. If there are fractions, multiply both sides by the lowest common multiple of the denominators, a positive number. When a fraction stands alone, as in $\frac{x}{3} + 2 \ge 5$ , you may move the number first and then multiply.

Step 2: Collect the terms in $x$ on one side and the numbers on the other, by adding or subtracting on both sides.

Step 3: Divide both sides by the coefficient of $x$ . If the coefficient is negative, reverse the sign in the same line.

Step 4: Check with one number from your answer, which must satisfy the inequality, and one number outside it, which must not.

Step 5: Show the solution on a number line.

Inequalities of the Form ax + b ≤ c

Example 1: Solve, check and draw

Solve $2x - 3 \ge 5$ , check your answer, and show the solution on a number line.

Solution

Add 3 to both sides: $2x \ge 8$ .

Divide both sides by 2, a positive number: $x \ge 4$ .

Check a number from the answer, $x = 5$ : $2(5) - 3 = 7$ , and $7 \ge 5$ is true.

Check a number outside it, $x = 3$ : $2(3) - 3 = 3$ , and $3 \ge 5$ is false.

$x \ge 4$

Example 2: Three more of the same form

Solve each inequality and show the solution on a number line.
(a) $5x - 7 > 18$
(b) $4x + 1 < 11$
(c) $3x + 10 < 4$

Solution

(a) Add 7 to both sides: $5x > 25$ . Divide both sides by 5: $x > 5$ .

(b) Subtract 1 from both sides: $4x < 10$ . Divide both sides by 4: $x < \frac{10}{4}$ , so $x < 2.5$ . The boundary value is not a whole number.

Check $x = 2$ : $4(2) + 1 = 9$ , and $9 < 11$ is true. Check $x = 3$ : $13 < 11$ is false.

Check $x = -3$ : $3(-3) + 10 = 1$ , and $1 < 4$ is true. Check $x = 0$ : $10 < 4$ is false.

(a) $x > 5$ (b) $x < 2.5$ (or $x < \frac{5}{2}$ ) (c) $x < -2$

When the Coefficient of x Is Negative

When the term in $x$ has a negative coefficient, as in $-3x + 4$ or $7 - 2x$ , the last step divides by a negative number, and the sign must be reversed in that step.

Example 3: A negative coefficient

Solve each inequality and show the solution on a number line.
(a) $-3x + 4 \le 19$
(b) $7 - 2x \ge 1$
(c) $-5x + 3 < 23$

Solution

(a) Subtract 4 from both sides: $-3x \le 15$ .

Divide both sides by $-3$ , which reverses the sign: $x \ge \frac{15}{-3}$ , so $x \ge -5$ .

Check $x = 0$ : $-3(0) + 4 = 4$ , and $4 \le 19$ is true. Check $x = -6$ : $-3(-6) + 4 = 22$ , and $22 \le 19$ is false.

(b) Subtract 7 from both sides: $-2x \ge -6$ .

Divide both sides by $-2$ , which reverses the sign: $x \le \frac{-6}{-2}$ , so $x \le 3$ .

Another way, with no reversal: add $2x$ to both sides, $7 \ge 1 + 2x$ ; subtract 1, $6 \ge 2x$ ; divide by 2, $3 \ge x$ , which is $x \le 3$ .

Check $x = 0$ : $7 \ge 1$ is true. Check $x = 4$ : $7 - 8 = -1$ , and $-1 \ge 1$ is false.

Divide both sides by $-5$ , which reverses the sign: $x > \frac{20}{-5}$ , so $x > -4$ .

Check $x = 0$ : $3 < 23$ is true. Check $x = -5$ : $-5(-5) + 3 = 28$ , and $28 < 23$ is false.

(a) $x \ge -5$ (b) $x \le 3$ (c) $x > -4$

The Unknown on Both Sides

When $x$ is on both sides, subtract one of the $x$ terms from both sides. Taking away the $x$ term with the smaller coefficient leaves a positive coefficient, so there is nothing to reverse. Taking away the one with the larger coefficient also works, as long as the sign is reversed when you divide.

Example 4: x on both sides

Solve each inequality and show the solution on a number line.
(a) $5x - 3 > 2x + 9$
(b) $2x + 7 \le 5x - 2$

Solution

(a) Subtract $2x$ from both sides: $3x - 3 > 9$ . Add 3 to both sides: $3x > 12$ . Divide both sides by 3: $x > 4$ .

Check $x = 5$ : the left side is $22$ and the right side is $19$ , and $22 > 19$ is true. Check $x = 4$ : both sides are 17, and $17 > 17$ is false, so 4 is not a solution, as the open circle shows.

(b) The $x$ term with the larger coefficient is on the right. Subtract $2x$ from both sides: $7 \le 3x - 2$ . Add 2 to both sides: $9 \le 3x$ . Divide both sides by 3: $3 \le x$ , which is $x \ge 3$ .

The other way: subtract $5x$ from both sides, $-3x + 7 \le -2$ ; subtract 7, $-3x \le -9$ ; divide by $-3$ , which reverses the sign, $x \ge 3$ . The answer is the same.

Check $x = 4$ : $15 \le 18$ is true. Check $x = 2$ : $11 \le 8$ is false.

(a) $x > 4$ (b) $x \ge 3$

Inequalities with Brackets

Example 5: Expanding the brackets first

Solve each inequality and show the solution on a number line.
(a) $3(x - 2) \le 9$
(b) $2(3 - x) > 4$
(c) $4(x + 1) - 3 \ge 2(x - 3)$

Solution

(a) Expand the brackets: $3x - 6 \le 9$ . Add 6 to both sides: $3x \le 15$ . Divide both sides by 3: $x \le 5$ .

Check $x = 5$ : $3(3) = 9$ , and $9 \le 9$ is true. Check $x = 6$ : $3(4) = 12$ , and $12 \le 9$ is false.

(b) Expand the brackets: $6 - 2x > 4$ . Subtract 6 from both sides: $-2x > -2$ . Divide both sides by $-2$ , which reverses the sign: $x < 1$ .

Check $x = 0$ : $2(3) = 6$ , and $6 > 4$ is true. Check $x = 2$ : $2(1) = 2$ , and $2 > 4$ is false.

Subtract $2x$ from both sides: $2x + 1 \ge -6$ . Subtract 1 from both sides: $2x \ge -7$ . Divide both sides by 2: $x \ge -3.5$ .

Check $x = 0$ : the left side is $1$ and the right side is $-6$ , and $1 \ge -6$ is true. Check $x = -4$ : the left side is $4(-3) - 3 = -15$ and the right side is $2(-7) = -14$ , and $-15 \ge -14$ is false.

(a) $x \le 5$ (b) $x < 1$ (c) $x \ge -3.5$

Inequalities with Fractions

To clear the fractions, multiply every term on both sides by the lowest common multiple of the denominators. It is positive, so the sign does not change.

Example 6: Clearing the fractions

Solve each inequality and show the solution on a number line.
(a) $\frac{x}{3} + 2 \ge 5$
(b) $\frac{2x - 1}{3} < 5$
(c) $\frac{3x}{4} - 1 \le \frac{x}{2}$

Solution

(a) Subtract 2 from both sides: $\frac{x}{3} \ge 3$ . Multiply both sides by 3: $x \ge 9$ .

Check $x = 12$ : $4 + 2 = 6$ , and $6 \ge 5$ is true. Check $x = 6$ : $2 + 2 = 4$ , and $4 \ge 5$ is false.

(b) Multiply both sides by 3. The whole of $2x - 1$ is divided by 3, so it is all multiplied back: $2x - 1 < 15$ . Add 1 to both sides: $2x < 16$ . Divide both sides by 2: $x < 8$ .

Check $x = 5$ : $\frac{9}{3} = 3$ , and $3 < 5$ is true. Check $x = 8$ : $\frac{15}{3} = 5$ , and $5 < 5$ is false.

(c) The lowest common multiple of 4 and 2 is 4. Multiply every term by 4: $3x - 4 \le 2x$ . Subtract $2x$ from both sides: $x - 4 \le 0$ . Add 4 to both sides: $x \le 4$ .

Check $x = 4$ : $3 - 1 = 2$ and $\frac{4}{2} = 2$ , and $2 \le 2$ is true. Check $x = 8$ : $6 - 1 = 5$ and $\frac{8}{2} = 4$ , and $5 \le 4$ is false.

(a) $x \ge 9$ (b) $x < 8$ (c) $x \le 4$

Common Mistakes

Reversing the sign when no negative number is divided by. In $3x + 10 < 4$ , subtracting 10 gives $3x < -6$ , and dividing by positive 3 gives $x < -2$ . Nothing is reversed.
Not reversing when the coefficient is negative. In $7 - 2x \ge 1$ , the last step divides by $-2$ , so $\ge$ becomes $\le$ : $x \le 3$ . Or collect $x$ on the side where its coefficient is positive.
Multiplying only part of a side. $\frac{2x - 1}{3} < 5$ gives $2x - 1 < 15$ . Every term on both sides is multiplied by 3.
Expanding a bracket with a minus sign wrongly. $2(3 - x)$ is $6 - 2x$ , not $6 - x$ or $6 + 2x$ .
Checking only the boundary value. It shows whether the boundary value is included, but not which way the solutions go. Check one number from the answer and one outside it as well.

Practice

Solve each inequality on paper, and show the solution on a number line, before you open its solution.

Question 1

Solve each inequality and show the solution on a number line.
(a) $2x + 5 > 17$
(b) $4x - 9 \le 7$

ShowHide solution

(a) Subtract 5 from both sides: $2x > 12$ . Divide both sides by 2: $x > 6$ .

(b) Add 9 to both sides: $4x \le 16$ . Divide both sides by 4: $x \le 4$ .

(a) $x > 6$ (b) $x \le 4$

Question 2

Solve each inequality and show the solution on a number line.
(a) $6x + 1 \ge 10$
(b) $2x + 9 < 3$

ShowHide solution

(a) Subtract 1 from both sides: $6x \ge 9$ . Divide both sides by 6: $x \ge \frac{9}{6}$ , so $x \ge 1.5$ .

(b) Subtract 9 from both sides: $2x < -6$ . Divide both sides by 2, a positive number: $x < -3$ .

(a) $x \ge 1.5$ (b) $x < -3$

Question 3

Solve each inequality and show the solution on a number line.
(a) $-4x + 1 < 13$
(b) $-2x - 7 \ge 5$

ShowHide solution

(a) Subtract 1 from both sides: $-4x < 12$ . Divide both sides by $-4$ , which reverses the sign: $x > -3$ .

(b) Add 7 to both sides: $-2x \ge 12$ . Divide both sides by $-2$ , which reverses the sign: $x \le -6$ .

(a) $x > -3$ (b) $x \le -6$

Question 4

Solve each inequality and show the solution on a number line.
(a) $10 - 3x > 4$
(b) $1 - 4x \le 7$
(c) $5 - 4x > -7$

ShowHide solution

(a) Subtract 10 from both sides: $-3x > -6$ . Divide both sides by $-3$ , which reverses the sign: $x < 2$ .

(b) Subtract 1 from both sides: $-4x \le 6$ . Divide both sides by $-4$ , which reverses the sign: $x \ge \frac{6}{-4}$ , so $x \ge -1.5$ .

(a) $x < 2$ (b) $x \ge -1.5$ (c) $x < 3$

Question 5

Solve each inequality and show the solution on a number line.
(a) $7x - 4 \ge 3x + 8$
(b) $x + 11 < 4x - 1$
(c) $3 - 2x \le x + 9$

ShowHide solution

(a) Subtract $3x$ from both sides: $4x - 4 \ge 8$ . Add 4: $4x \ge 12$ . Divide by 4: $x \ge 3$ .

(b) Subtract $x$ from both sides: $11 < 3x - 1$ . Add 1: $12 < 3x$ . Divide by 3: $4 < x$ , which is $x > 4$ .

Check $x = 0$ : $3 \le 9$ is true. Check $x = -3$ : $9 \le 6$ is false.

(a) $x \ge 3$ (b) $x > 4$ (c) $x \ge -2$

Question 6

Solve each inequality and show the solution on a number line.
(a) $4(x + 3) < 20$
(b) $5(2 - x) \ge 15$
(c) $3(2x - 1) > 2(x + 5)$

ShowHide solution

(a) Expand: $4x + 12 < 20$ . Subtract 12: $4x < 8$ . Divide by 4: $x < 2$ .

(b) Expand: $10 - 5x \ge 15$ . Subtract 10: $-5x \ge 5$ . Divide by $-5$ , which reverses the sign: $x \le -1$ .

Check $x = 4$ : $3(7) = 21$ and $2(9) = 18$ , and $21 > 18$ is true. Check $x = 3$ : $3(5) = 15$ and $2(8) = 16$ , and $15 > 16$ is false.

(a) $x < 2$ (b) $x \le -1$ (c) $x > 3.25$ (or $x > \frac{13}{4}$ )

Question 7

Solve each inequality and show the solution on a number line.
(a) $\frac{x}{5} \le 4$
(b) $\frac{x}{4} - 3 > 1$
(c) $\frac{3x + 2}{4} \ge 5$
(d) $1 - \frac{x}{2} < 4$

ShowHide solution

(a) Multiply both sides by 5: $x \le 20$ .

(b) Add 3 to both sides: $\frac{x}{4} > 4$ . Multiply both sides by 4: $x > 16$ .

(d) Subtract 1 from both sides: $-\frac{x}{2} < 3$ . Multiply both sides by $-2$ , which reverses the sign: $x > -6$ .

Check $x = 0$ : $1 < 4$ is true. Check $x = -8$ : $1 + 4 = 5$ , and $5 < 4$ is false.

(a) $x \le 20$ (b) $x > 16$ (c) $x \ge 6$ (d) $x > -6$

Question 8

(a) Solve $3(x + 2) - x \le 14$ .
(b) Check your answer with $x = 4$ , $x = 0$ and $x = 5$ .
(c) Show the solution on a number line.

ShowHide solution

(a) Expand: $3x + 6 - x \le 14$ , so $2x + 6 \le 14$ . Subtract 6: $2x \le 8$ . Divide by 2: $x \le 4$ .

(b) $x = 4$ : $3(6) - 4 = 14$ , and $14 \le 14$ is true. $x = 0$ : $3(2) - 0 = 6$ , and $6 \le 14$ is true. $x = 5$ : $3(7) - 5 = 16$ , and $16 \le 14$ is false. These agree with $x \le 4$ : 4 and 0 are in the answer, and 5 is not.

(a) $x \le 4$ (b) 4 and 0 satisfy the inequality; 5 does not.

Question 9

Amir solved $7 - 3x < 1$ and wrote the answer $x < 2$ .
(a) Test his answer with $x = 0$ .
(b) Solve the inequality correctly and show the solution on a number line.

ShowHide solution

(a) $x = 0$ satisfies $x < 2$ . But $7 - 3(0) = 7$ , and $7 < 1$ is false. So $x < 2$ is not the answer.

(b) Subtract 7 from both sides: $-3x < -6$ . Divide both sides by $-3$ , which reverses the sign: $x > 2$ . Amir did not reverse the sign.

Check $x = 3$ : $7 - 9 = -2$ , and $-2 < 1$ is true.

(a) $x = 0$ does not satisfy $7 - 3x < 1$ . (b) $x > 2$

Question 10

Three times a number $n$ , decreased by 8, is at least 13.
(a) Write an inequality in $n$ .
(b) Solve it and show the solution on a number line.

ShowHide solution

(a) Three times the number, decreased by 8, is $3n - 8$ . "At least 13" allows 13 and anything more.

(b) Add 8 to both sides: $3n \ge 21$ . Divide by 3: $n \ge 7$ .

(a) $3n - 8 \ge 13$ (b) $n \ge 7$ : the number is 7 or more.

Question 11

When a number $n$ is doubled and 5 is added, the result is less than the number increased by 9.
(a) Write an inequality in $n$ .
(b) Solve it and show the solution on a number line.

ShowHide solution

(a) Doubled and 5 added: $2n + 5$ . Increased by 9: $n + 9$ .

(b) Subtract $n$ from both sides: $n + 5 < 9$ . Subtract 5: $n < 4$ .

Check $n = 3$ : $11 < 12$ is true. Check $n = 4$ : $13 < 13$ is false.

(a) $2n + 5 < n + 9$ (b) $n < 4$ : the number is less than 4.

Question 12

A taxi fare is $4 plus $0.80 for every kilometre. Jun Hao has $20 and wants the fare to be at most $20. Let the distance be $d$ km.
(a) Write an inequality in $d$ .
(b) Solve it and show the solution on a number line.

ShowHide solution

(a) The fare for $d$ km is $4 + 0.8d$ dollars, and "at most 20" allows 20 and anything less.

(b) Subtract 4 from both sides: $0.8d \le 16$ . Divide by 0.8: $d \le 20$ .

Check $d = 20$ : $4 + 16 = 20$ , and $20 \le 20$ is true. Check $d = 21$ : $4 + 16.8 = 20.8$ , and $20.8 \le 20$ is false.

(a) $4 + 0.8d \le 20$ (b) $d \le 20$ : Jun Hao can travel at most 20 km.

Key Points

Solve a linear inequality as you would a linear equation: expand the brackets, clear the fractions, collect the terms in $x$ on one side, then divide by the coefficient of $x$ .
Only multiplying or dividing by a negative number reverses the sign, and the reversed sign goes in that same line: $-2x \ge -6$ gives $x \le 3$ .
To avoid a reversal, collect $x$ on the side where its coefficient is positive: $7 - 2x \ge 1$ gives $6 \ge 2x$ , so $x \le 3$ .
Multiply every term by the lowest common multiple of the denominators to clear fractions.
Check with one number from the answer and one outside it, and show every answer on a number line.