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Linear Inequalities

In this section we learn how to solve linear inequalities. These are inequalities involving an unknown $$x$$, looking like either of the following: $ax \leq b, \quad ax+b > d$ or $ax+b \geq cx, \quad ax+b < cx +d$ Those are all typical linear inequalities, that we'll know how to solve by the end of this section.

Method - Solving Inequalites

To solve a linear equality looking like $$ax \leq b$$ or even $$ax +b > cx + d$$, we follow the same steps as we would for solving the linear equations $$ax = b$$ and $$ax +b = cx + d$$ with one very important difference:

When we multiply, or divide, both sides of an inequality by a negative number: the inequality symbol is reversed.

In other words, when we multiply, or divide, both sides of an inequality by a negative number:

$$>$$ becomes $$<$$

$$\geq$$ becomes $$\leq$$

$$<$$ becomes $$>$$

$$\leq$$ becomes $$\geq$$

This is further explained and some examples are worked through in Tutorial 1. If you haven't done so already, watch it now.

One-Step Linear Inequalities

Now that we've learnt the rules for solving linear inequalities, we can put our knowledge into practice.
We start by working our way through a few one-step inequalities. If you're having any trouble solving these, make sure to watch Tutorial 2 and to carefully read through the "Answers With Working" for each question.

Exercise 1

Solve each of the following inequalities and illustrate your answer on the number line:

1. $$2x \leq 6$$

2. $$-3x < 9$$

3. $$\frac{x}{2} \geq 2$$

4. $$x-2>0$$

5. $$\frac{x}{-4} \leq -1$$

1. $$x\leq 3$$
2. $$x> -3$$
3. $$x \geq 4$$
4. $$x>2$$
5. $$x \geq 4$$

1. We solve $$2x \leq 6$$ as follows: $2x \leq 6$ Divide both sides by $$2$$: $\frac{2x}{2} \leq \frac{6}{2}$ $x \leq 3$ This result is illustrated on the number line as follows:
Notice that the dot above the $$3$$ is filled-in to indicate that the value $$3$$ is included in the solution.

2. We solve $$-3x < 9$$ as follows: $-3x < 9$ Divide both sides by $$-3$$ and reverse the inequality symbol: $\frac{-3x}{-3} > \frac{9}{-3}$ $x > -3$ This result is illustrated on the number line as follows:
Notice that the dot above the $$3$$ is empty to indicate that the value $$-3$$ is not included in the solution.

3. We solve $$\frac{x}{2} \geq 2$$ as follows: $\frac{x}{2} \geq 2$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$: $2\times \frac{x}{2} \geq 2\times 2$ $x \geq 4$ This result is illustrated on the number line as follows:
Notice that the dot above the $$4$$ is filled-in to indicate that the value $$4$$ is included in the solution.

4. We solve $$x-2>0$$ as follows: $x-2 > 0$ Add $$2$$ to both sides of the inequality to get rid of the $$2$$ that is being subtracted from $$x$$: $x-2+2 > 0 + 2$ $x > 2$ This result is illustrated on the number line as follows:
Notice that the dot above the $$2$$ is empty to indicate that $$2$$ isn't included in the solution.

5. We solve $$\frac{x}{-4} \leq -1$$ as follows: $\frac{x}{-4} \leq -1$ Multiply both sides by $$-4$$ to get rid of the $$-4$$ that is dividing $$x$$ and reverse the inequality symbol: $-4 \times \frac{x}{-4} \geq -4 \times (-1)$ $x \geq 4$ This result is illustrated on the number line as follows:
Notice that the dot above the $$4$$ is filled-in to indicate that $$4$$ is included in the solution.

Two-Step Linear Inequalities

Now that we've seen how to solve simple linear inequalities, that can be solved in one step, we move-on to solving inequalities that require two steps. If you're having any trouble solving these, make sure to watch Tutorial 2 and to carefully read through the "Answers With Working" for each question.

Exercise 2

Solve each of the following inequalities annd illustrate your answers on the number line:

1. $$4x+1 > 13$$

2. $$2 - 5x \geq 22$$

3. $$3 < 17 - 7x$$

4. $$12 + 6x > 24$$

5. $$1 - \frac{x}{2} \leq 4$$

1. $$x > 3$$
2. $$x \leq -4$$
3. $$x < 2$$
4. $$x > 2$$
5. $$x \geq -6$$

1. We solve $$4x + 1 > 13$$ as follows: $4x + 1 > 13$ Subtract $$1$$ from both sides, to get rid of the $$1$$ that's being added to $$4x$$: $4x + 1 -1 > 13 - 1$ $4x > 12$ Divide both sides by $$4$$, to get rid of the $$4$$ that is multiplying the $$x$$: $\frac{4x}{4} > \frac{12}{4}$ $x > 3$ This result is illustrated here:
Notice that the dot above the $$3$$ is empty to indicate that $$3$$ is not included in the solution.

2. We solve $$2 - 5x \geq 22$$ as follows: $2 - 5x \geq 22$ Subtract $$2$$ from both sides, to get rid of the positive $$2$$ on the left hand side: $2 - 5x -2\geq 22-2$ $- 5x \geq 20$ Divide both sides by $$-5$$, and reverse the inequality symbol, to get rid of the $$-5$$ that is multiplying the $$x$$: $\frac{-5x}{-5} \leq \frac{20}{-5}$ $x \leq -4$ This result is illustrated here:
Notice that the dot above $$-4$$ is filled-in to indicate that $$-4$$ is included in the solution.

3. We solve $$3< 17 - 7x$$ as follows: $3 < 17 - 7x$ Subtract $$17$$ from both sides to get rid of the the positive $$17$$ on the right hand side: $3 -17 < 17 - 7x - 17$ $-14 < -7x$ Divide noth sides by $$-7$$, and reverse the inequality symbol, to get rid of the $$-7$$ multiplying the $$x$$: $\frac{-14}{-7} > \frac{-7x}{-7}$ $2 > x$ This result "$$x$$ is less than $$2$$" can be re-written as: $x < 2$ We illustrate this as follows:

4. We solve $$12 + 6x > 24$$ as follows: $12 + 6x > 24$ Subtract $$12$$ from both sides, to get rid of the $$12$$ that's being added to $$6x$$: $12 + 6x -12 > 24 -12$ $6x > 12$ Divide both sides by $$6$$ to get rid of the $$6$$ that's multiplying the $$x$$: $\frac{6x}{6} > \frac{12}{6}$ $x > 2$ This result is illustrated here:
Notice that the dot above the $$2$$ is empty to indicate that $$2$$ isn't included in the solution.

5. We solve $$1 - \frac{x}{2} \leq 4$$ as follows: $1 - \frac{x}{2} \leq 4$ Subtract $$1$$ from both sides to get rid of the positive $$1$$ on the left hand side: $1 - \frac{x}{2} -1 \leq 4 - 1$ $-\frac{x}{2} \leq 3$ Multiply both sides by $$-2$$, and change the inequality symbol, to get rid of the $$-2$$ dividing $$x$$ (make sure to read $$\text{Note}^{(1)}$$, further down, to fully undertand this step): $-2\times \begin{pmatrix} -\frac{x}{2} \end{pmatrix} \geq -2\times 3$ $x \geq -6$ This result is illustrated here:
Notice that the dot above $$-6$$ is filled-in to indicate that $$-6$$ is included in the solution.

$$\text{Note}^{(1)}$$: whether we write $$-\frac{x}{2}$$, $$\frac{-x}{2}$$, or $$\frac{x}{-2}$$ these are all equivalent notations and can be interchanged when convenient to do so.

Three-Step Linear Inequalities

We now move-on to linear inequalities that require three steps to be solved.
These usually have an $$x$$ term on each side of the inequality. If you're having any trouble solving these, make sure you have watched Tutorial 3 and to carefully read through the "answers with working".

Exercise 3

Solve each of the following inequalities and illustarate your answer on the number line:

1. $$3x+6 < x$$

2. $$2x + 3 \geq 3x$$

3. $$\frac{2x}{3} \leq x -2$$

4. $$4x - 5 > 2x -9$$

5. $$20 + 3x \geq 4 + 7x$$

1. $$x < -3$$
2. $$x \leq 3$$
3. $$x \geq 6$$
4. $$x > -2$$
5. $$x \leq 4$$

1. We solve $$3x + 6 < x$$ as follows: $3x + 6 < x$ Subtract $$6$$ from both sides to get rid of the $$6$$ being added to $$3x$$: $3x + 6 - 6 < x - 6$ $3x < x - 6$ Subtract $$x$$ from both sides to gather all the $$x$$ terms on the left hand side: $3x - x < x - 6 - x$ $2x < -6$ Divide both sides by $$2$$ to get rid of the $$2$$ multiplying the $$x$$: $\frac{2x}{2} < \frac{-6}{2}$ $x < -3$ This result is illustrated here:
Notice that that dot above the $$-3$$ is empty to indicate that $$-3$$ is not included in the solution.

2. We solve $$2x + 3 \geq 3x$$ as follows: $2x + 3 \geq 3x$ Subtract $$3$$ from both sides to get rid of the $$3$$ being added to $$2x$$: $2x + 3 -3 \geq 3x - 3$ $2x \geq 3x -3$ Subtract $$3x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x -3x\geq 3x -3 -3x$ $-x \geq -3$ Multiply both sides by $$-1$$, and reverse the inequality symbol, to get rid of the negative on the $$x$$: $-1\times (-x) \leq -1 \times (-3)$ $x \leq 3$ This result is illustrated here:
Notice that the dot above the $$3$$ is filled-in to indicate that $$3$$ is included in the solution.

3. We solve $$\frac{2x}{3} \leq x - 2$$ as follows: $\frac{2x}{3} \leq x - 2$ Multiply both sides by $$3$$ to get rid of the $$3$$ that's dividing the $$x$$ on the left hand side: $3\times \frac{2x}{3} \leq 3\begin{pmatrix} x - 2 \end{pmatrix}$ $2x \leq 3\times x - 3\times 2$ $2x \leq 3x - 3$ Subtract $$3x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x - 3x \leq 3x - 6 - 3x$ $-x \leq -3$ Multiply both sides by $$-1$$, and reverse the inequality symbol, to get rid of the negative on the $$x$$: $-1\times (-x) \geq -1 \times (-6)$ $x \geq 6$ This result is illustrated here:
Notice that the dot, above the $$6$$, is filled-in to indicate that $$6$$ is included in the solution.

4. We solve $$4x - 5 > 2x - 9$$ as follows: $4x - 5 > 2x - 9$ Add $$5$$ to both sides to get rid of the $$5$$ that is being subtracted from $$4x$$: $4x - 5 + 5 > 2x - 9 + 5$ $4x > 2x - 4$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $4x -2x > 2x - 4 - 2x$ $2x > -4$ Divide both sides by $$2$$ to get rid of the $$2$$ that's multiplying $$x$$: $\frac{2x}{2} > \frac{-4}{2}$ $x > -2$ This result is illustrated here:
Notice that the dot above the $$-2$$ is empty to indicate that $$-2$$ is not included in the solution.

5. We solve $$20 + 3x \geq 4 + 7x$$ as follows: $20 + 3x \geq 4 + 7x$ Subtract $$20$$ from both sides to get rid of the $$20$$ that's being added to $$3x$$ on the left hand side: $20 + 3x - 20 \geq 4 + 7x - 20$ $3x \geq 7x - 16$ Subtract $$7x$$ from both sides to gather all the $$x$$ terms on the left hand side: $3x -7x \geq 7x - 16 -7x$ $-4x \geq -16$ Divide both sides by $$-4$$, and reverse the inequality symbol, to get rid of the $$-4$$ multiplying the $$x$$: $\frac{-4x}{-4} \leq \frac{-16}{-4}$ $x \leq 4$ This result is illustrated here:
Notice that the dot above the $$4$$ is filled-in to indicate that $$4$$ is included in the solution.

Four-Step, or more, Linear Inequalities

We now move-on to more complicated linear inequalities that require four, or more, steps to be solved.
These usually involve fractions. If you're having any trouble solving these, make sure you have watched Tutorial 3 and to carefully read through the "answers with working".

Exercise 4

Solve each of the following inequalities and illustrate your answer on the number line:

1. $$\frac{x}{2} + 1 \leq 2x + 4$$

2. $$5x - 1 > 2x +5$$

3. $$\frac{2x}{3} - 2 \geq 2x + 2$$

4. $$-2 - \frac{x}{2} > \frac{x}{3} + 8$$

5. $$\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$$

1. $$x \geq -2$$
2. $$x > 2$$
3. $$x \leq -3$$
4. $$x < -12$$
5. $$x \geq 6$$

1. We solve $$\frac{x}{2} + 1 \leq 2x +4$$ as follows: $\frac{x}{2} + 1 \leq 2x +4$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$: $2\begin{pmatrix} \frac{x}{2} + 1 \end{pmatrix} \leq 2\begin{pmatrix} 2x +4 \end{pmatrix}$ $2\times \frac{x}{2} + 2\times 1 \leq 2\times 2x + 2\times 4$ $x + 2 \leq 4x + 8$ Subtract $$2$$ from both sides to get rid pf the $$2$$ that is being added to $$x$$ on the left hand side: $x + 2 -2 \leq 4x + 8 -2$ $x \leq 4x + 6$ Subtract $$4x$$ from both sides to gather all the $$x$$ terms on the left hand side: $x -4x \leq 4x + 6 -4x$ $-3x \leq 6$ Divide both sides by $$-3$$, and reverse the inequality symbol, to get rid of the $$-3$$ that's multiplying the $$x$$: $\frac{-3x}{-3} \geq \frac{6}{-3}$ $x \geq -2$ This result is illustrated here:
Notice that the dot above $$-2$$ is filled-in to indicate that $$-2$$ is included in the solution.

2. We solve $$5x - 1 > 2x + 5$$ as follows: $5x - 1 > 2x + 5$ Add $$1$$ to both sides to get rid of the $$1$$ that is being subtracted from $$5x$$: $5x - 1 + 1 > 2x + 5 + 1$ $5x > 2x + 6$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $5x - 2x > 2x + 6 - 2x$ $3x > 6$ Divide both sides by $$3$$ to get rid of the $$3$$ multiplying the $$x$$ on the left hand side: $\frac{3x}{3} > \frac{6}{3}$ $x > 2$ This result is illustrated here:
Notice that the dot above $$2$$ is empty to indicate that $$2$$ is not included in the solution.
3. We solve $$\frac{2x}{3} - 2 \geq 2x + 2$$ as follows: $\frac{2x}{3} - 2 \geq 2x + 2$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is dividing $$x$$ on the left hand side: $3 \begin{pmatrix} \frac{2x}{3} - 2 \end{pmatrix} \geq 3 \begin{pmatrix} 2x + 2 \end{pmatrix}$ $3\times \frac{2x}{3} - 3 \times 2 \geq 3 \times 2x + 3 \times 2$ $2x - 6 \geq 6x + 6$ Add $$6$$ to both sides to get rid of the $$6$$ that's being subtracted from $$2x$$: $2x - 6 + 6 \geq 6x + 6 + 6$ $2x \geq 6x + 12$ Subtract $$6x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x - 6x \geq 6x + 12 - 6x$ $-4x \geq 12$ Divide both sides by $$-4$$, and reverse the inequality symbol, to get rid of the $$-4$$ that's multiplying $$x$$: $\frac{-4x}{-4} \leq \frac{12}{-4}$ $x \leq -3$ This result is illustrated here:
Notice that the dot above $$-3$$ is filled-in to indicate that $$-3$$ is included in the solution.

4. We solve $$-2 - \frac{x}{2} > \frac{x}{3} + 8$$ as follows: $-2 - \frac{x}{2} > \frac{x}{3} + 8$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$ on the left hand side: $2\begin{pmatrix} -2 - \frac{x}{2} \end{pmatrix} > 2 \begin{pmatrix} \frac{x}{3} + 8 \end{pmatrix}$ $2\times (-2) - 2 \times \frac{x}{2} > 2 \times \frac{x}{3} + 2 \times 8$ $-4 - x > \frac{2x}{3} + 16$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is diving $$x$$ on the right hand side: $3\begin{pmatrix} -4 - x \end{pmatrix} > 3\begin{pmatrix} \frac{2x}{3} + 16 \end{pmatrix}$ $3\times (-4) - 3 \times x > 3 \times \frac{2x}{3} + 3\times 16$ $-12 -3x > 2x + 48$ Add $$12$$ to both sides to get rid of the $$-12$$ on the left hand side: $-12 - 3x + 12 > 2x + 48 + 12$ $-3x > 2x + 60$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $-3x - 2x > 2x + 60 - 2x$ $-5x > 60$ Divide both sides by $$-5$$, and reverse the inequality symbol, to get rid of the $$-5$$ that's multiplying the $$x$$: $\frac{-5x}{-5} < \frac{60}{-5}$ $x < -12$ This result is illustrated here:
Notice that the dot above $$-12$$ is empty to indicate that $$-12$$ is not included in the solution.

5. We solve $$\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$$ as follows: $\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$ Multiply both sides by $$2$$ to get rid of the $$2$$ dividing the $$x$$ on the left hand side: $2\begin{pmatrix} \frac{3x}{2} - 1 \end{pmatrix} \geq 2 \begin{pmatrix} \frac{x}{3} + 6 \end{pmatrix}$ $2 \times \frac{3x}{2} -2 \times 1 \geq 2 \times \frac{x}{3} + 2 \times 6$ $3x - 2 \geq \frac{2x}{3} + 12$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is dividing the $$x$$ on the right hand side: $3\begin{pmatrix} 3x - 2 \end{pmatrix} \geq 3 \begin{pmatrix} \frac{2x}{3} + 12 \end{pmatrix}$ $3 \times 3x - 3 \times 2 \geq 3 \times \frac{2x}{3} + 3 \times 12$ $9x - 6 \geq 2x + 36$ Add $$6$$ to both sides to get rid of the $$6$$ that is being subtracted from $$9x$$ on the left hand side: $9x - 6 + 6 \geq 2x + 36 + 6$ $9x \geq 2x + 42$ Subtract $$2x$$ from both sides to gather all of the $$x$$ terms on the left hand side: $9x - 2x \geq 2x + 42 - 2x$ $7x \geq 42$ Divide both sides by $$7$$ to get rid of the $$7$$ multiplying the $$x$$: $\frac{7x}{7} \geq \frac{42}{7}$ $x \geq 6$ This result is illustrated here:
Notice that the dot above $$6$$ is filled-in to indicate that $$6$$ is included in the solution.