National 5 Maths
Linear Inequalities

The letter \(x\) appears only once in this very simple inequality.

So we just perform the necessary operations to make \(x\) the subject.

First, we will subtract \(2\) from both sides. Then we will divide both sides by \(4\small.\)

Because neither step involves multiplying or dividing by a negative number, the \(\lt\) sign does not change.

$$ \begin{gather} 4x+2\lt 9\\[6pt] 4x\lt 9-2\\[6pt] 4x\lt 7\\[6pt] x\lt\small\frac{7}{4} \end{gather} $$

Note: This should be left as an improper fraction. It is usually a good idea to avoid mixed numbers and decimals.

Video solution by Clelland Maths 

In this example, the letter term \(2z\) is on the right hand side. It has a positive coefficient: \(2\small.\)

One way to handle this is to switch sides at the beginning of our working. The \(\gt\) sign becomes a \(\lt\) sign.

$$ \begin{gather} 4\gt 2z-5\\[6pt] 2z-5\lt 4\\[6pt] 2z\lt 4+5\\[6pt] 2z\lt 9\\[6pt] z\lt\small\frac{9}{2} \end{gather} $$

Of course, we don't have to switch sides at the beginning. Another way of solving this is to subtract \(2z\) from both sides. We just need to remember to switch the inequality sign when dividing by \(-2\small.\) Here's how that looks:

$$ \begin{gather} 4\gt 2z-5\\[6pt] 4-2z\gt -5\\[6pt] -2z\gt -5-4\\[6pt] -2z\gt -9\\[6pt] z\lt\small\frac{-9}{-2}\\[6pt] z\lt\small\frac{9}{2} \end{gather} $$

Both methods give the correct solution, but the first method is definitely tidier.

Video solution by Clelland Maths 

In this example, the letter is on the right hand side and it has a negative coefficient.

Probably the most efficient way to start this solution is to add \(3x\) to both sides:

$$ \begin{gather} 10\lt 7-3x\\[6pt] 10+3x\lt 7\\[6pt] 3x\lt 7-10\\[6pt] 3x\lt -3\\[6pt] x\lt\small\frac{-3}{3}\\[6pt] x\lt -1\\[6pt] \end{gather} $$

Note: Adding \(3x\) to each side made this solution simpler than it would otherwise have been. We didn't have to divide by a negative number or switch sides, so the \(\lt\) sign didn't change to \(\gt\small.\)

Video solution by Clelland Maths 

This is our first example with letter terms on both sides.

First, we subtract \(5a\) from both sides and add \(2\) to both sides. We can do this in two separate steps, or combined into one step, as below.

$$ \begin{gather} 3a-2\geqslant 4+5a\\[8pt] 3a-5a\geqslant 4+2\\[8pt] -2a\geqslant 6\\[6pt] a \leqslant\small\frac{6}{-2}\\[6pt] a\leqslant -3 \end{gather} $$

Note 1: We needed to divide by a negative number. That's why the \(\geqslant\) sign changed to \(\leqslant\small.\)

Note 2: At time of writing, the exam board has never used \(\geqslant\) or \(\leqslant\) in a question. They have only used \(\gt\) or \(\lt\small.\) However, there is nothing in the specification  to stop them from using \(\geqslant\) or \(\leqslant\) in future.

Video solution by Clelland Maths 

The strategy here is to expand the bracket, collect like terms, multiply the whole equation through by 3 to get rid of the fraction, separate the letter and number terms onto opposite sides, and then solve.

$$ \begin{gather} \small\frac{2}{3}\normalsize x+4\gt 10-2(x-1)\\[8pt] \small\frac{2}{3}\normalsize x+4\gt 10-2x+2\\[8pt] \small\frac{2}{3}\normalsize x+4\gt 12-2x\\[10pt] 2x+12\gt 36-6x\\[10pt] 2x+6x\gt 36-12\\[10pt] 8x\gt 24\\[10pt] x\gt 3 \end{gather} $$

Note: The \(\gt\) sign didn't change in this example, because we divided by a positive number.

Video solution by Clelland Maths 

The negative sign on the right hand side needs to be handled correctly. Unlike with equations, it isn't a good idea to try and cross-multiply an inequality.

To get rid of the fractions, we use the lowest common multiple of 5 and 4, which is 20.

We have a choice: we can either use 20 as a common denominator or multiply through by 20. Both methods work well. We will use 20 as a common denominator below. You might like to try it the other way, as an exercise.

$$ \begin{gather} \small\frac{x+2}{5}\normalsize \gt -\small\frac{3x}{4}\normalsize \\[10pt] \small\frac{4(x+2)}{20}\normalsize \gt -\small\frac{15x}{20}\normalsize \\[10pt] 4(x+2)\gt -15x\\[10pt] 4x+8\gt -15x \\[10pt] 4x+15x\gt -8\\[8pt] 19x\gt -8\\[8pt] x\gt -\small\frac{8}{19} \end{gather} $$

Video solution by Clelland Maths 

The lowest common multiple of 4 and 6 is 12. We use the 12 to get rid of the fractions.

This can be done in two different ways: either we make a common denominator of 12 or we multiply the whole inequality through by 12 at the start.

In the solution below, we will start by multiplying through by 12. Why not try it the other way, as an exercise?

$$ \begin{gather} \small\frac{3}{4}\normalsize (x-5)-\small\frac{1}{6}\normalsize (x+2)\gt 7 \\[8pt] \small\frac{36}{4}\normalsize (x-5)-\small\frac{12}{6}\normalsize (x+2)\gt 84 \\[8pt] 9(x-5)-2(x+2)\gt 84 \\[10pt] 9x-45-2x-4\gt 84 \\[10pt] 9x-2x\gt 84+45+4 \\[10pt] 7x\gt 133 \\[9pt] x\gt\small\frac{133}{7} \\[9pt] x\gt 19 \end{gather} $$

Video solution by Clelland Maths 

The lowest common multiple of 6 and 8 is 24. We use the 24 to get rid of the fractions.

This can be done in two different ways: either we make a common denominator of 24 or we multiply the whole inequality through by 24 at the start.

In the solution below, we will start by multiplying through by 24. Why not try it the other way, as an exercise?

$$ \begin{gather} \small\frac{r}{6}\normalsize\leqslant\small\frac{1}{8}\normalsize-r \\[10pt] \small\frac{24r}{6}\normalsize\leqslant\small\frac{24}{8}\normalsize-24r \\[10pt] 4r\leqslant 3-24r \\[10pt] 4r+24r\leqslant 3 \\[10pt] 28r\leqslant 3 \\[10pt] r\leqslant \small\frac{3}{28}\normalsize \\[10pt] \end{gather} $$

Video solution by Clelland Maths 

The strategy here is to expand the bracket, collect like terms and then solve in the usual way:

A common mistake on this question is to break the BODMAS/BIDMAS rules by doing \(11-2\) first. However, the correct way is to multiply the terms in the bracket by \(-2\) first, and only then to subtract that from \(11\small.\)

$$ \begin{gather} 11-2(1+3x)\lt 39\\[6pt] 11-2-6x\lt 39\\[6pt] 9-6x\lt 39\\[6pt] -6x\lt 39-9\\[6pt] -6x\lt 30\\[6pt] x\gt\small\frac{30}{-6}\\[6pt] x\gt -5 \end{gather} $$

Note: Dividing by \(-6\) in the last step switched the \(\lt\) sign to \(\gt\small.\)

Video solution by Clelland Maths 

The method here is to expand \(3(x-2)\small,\) collect like terms and then rearrange:

$$ \begin{gather} 19+x\gt 15+3(x-2)\\[8pt] 19+x\gt 15+3x-6\\[8pt] 19+x\gt 3x+9\\[8pt] x-3x\gt 9-19\\[6pt] -2x\gt -10\\[6pt] x\lt\small\frac{-10}{-2}\\[6pt] x\lt 5\\[6pt] \end{gather} $$

Note: Dividing by \(-2\) in the last step switched the \(\gt\) sign to \(\lt\small.\)

Video solution by Clelland Maths 

The strategy here is to expand \(6(x-1)\small,\) collect like terms and then solve in the usual way:

$$ \begin{gather} 3x\lt 6(x-1)-12\\[8pt] 3x\lt 6x-6-12\\[8pt] 3x\lt 6x-18\\[8pt] 3x-6x\lt -18\\[6pt] -3x\lt -18\\[6pt] x\gt\small\frac{-18}{-3}\normalsize\\[6pt] x\gt 6 \end{gather} $$

Note: Dividing by \(-3\) in the last step switched the \(\lt\) sign to \(\gt\small.\)

Video solution by Clelland Maths 

Note that the minus sign before \((x+4)\) means that we are subtracting both the \(x\) and the \(+4\small.\)

$$ \begin{gather} 1-(x+4)\gt 2x\\[8pt] 1-x-4\gt 2x\\[8pt] -x-3\gt 2x\\[8pt] -x-2x\gt 3\\[6pt] -3x\gt 3\\[6pt] x\lt\small\frac{3}{-3}\\[6pt] x\lt -1 \end{gather} $$

Note: Dividing by \(-3\) in the last step switched the \(\gt\) sign to \(\lt\small.\)

Video solution by Clelland Maths 

We need to multiply through by the lowest common multiple of 3 and 5, which is 15.

This can be done in two different ways: either by making a common denominator of 15 or just multiplying through by 15 at the start.

In the solution below, we will make a common denominator of 15. Why not try it the other way, as an exercise?

$$ \begin{gather} \small\frac{x\,+\,1}{3}\normalsize -2\gt\small\frac{3x}{5}\normalsize\\[8pt] \small\frac{5(x\,+\,1)}{15}\normalsize -\small\frac{30}{15}\normalsize\gt\small\frac{3(3x)}{15}\normalsize\\[8pt] 5(x+1)-30\gt 9x\\[8pt] 5x+5-30\gt 9x\\[8pt] 5x-25\gt 9x\\[6pt] 5x-9x\gt 25\\[6pt] -4x\gt 25\\[6pt] x\lt -\small\frac{25}{4} \end{gather} $$

Note: Dividing by \(-4\) in the last step switched the \(\gt\) sign to \(\lt\small.\)

Video solution by Clelland Maths 

The method here is to multiply out the bracket, collect like terms, and then solve as usual.

$$ \begin{gather} 5(x-2)+4\lt 7x+8\\[8pt] 5x-10+4\lt 7x+8\\[8pt] 5x-6\lt 7x+8\\[8pt] 5x-7x\lt 8+6\\[8pt] -2x\lt 14\\[8pt] x\gt\small\frac{14}{-2}\\[8pt] x\gt -7\\[6pt] \end{gather} $$

Note: Dividing by \(-2\) in the last step switched the \(\lt\) sign to \(\gt\small.\)

Video solution by Clelland Maths