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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.

Tutorial

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. \(\frac{x}{-4} \leq -1\)

Answers Without Working

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

Answers With Working

  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. \(1 - \frac{x}{2} \leq 4 \)

Answers Without Working

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

Answers With Working

  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. \(20 + 3x \geq 4 + 7x \)

Answers Without Working

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

Answers With Working

  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. \(\frac{3x}{2} - 1 \geq \frac{x}{3} + 6 \)

Answers Without Working

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

Answers With Working

  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.