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

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

Note that this should be left as an improper (top-heavy) fraction. Mixed numbers and decimals should usually be avoided.

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

One way to handle this situation is to switch sides at the beginning of our working.

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

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

Probably the most efficient way to handle this situation is to add \(3x\) to both sides.

$$ \begin{eqnarray} 10 &\lt& 7-3x\\[6pt] 10+3x &\lt& 7\\[6pt] 3x &\lt& 7-10\\[6pt] 3x &\lt& -\!3\\[6pt] x &\lt& -\!1\\[6pt] \end{eqnarray} $$

First, we subtract \(5a\) from both sides and add \(2\) to both sides. This gives us only \(a\) terms on the left and only constant terms on the right.

$$ \begin{eqnarray} 3a-2 &\ge& 4+5a\\[6pt] 3a-5a &\ge& 4+2\\[6pt] -2a &\ge& 6\\[6pt] a &\le& \small\frac{6}{-2}\normalsize\\[6pt] a &\le& -\!3 \end{eqnarray} $$

Note that the last step was to divide by a negative number. That's why the ≥ sign changed to ≤.

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.

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

Note that the 'greater than' sign didn't change in this example, because we divided by a positive number.

The negative sign on the right hand side needs to be handled correctly. We do not recommend trying to cross-multiply with inequalities.

Our first job is to get rid of the fractions. To do this, we use the lowest common multiple of 5 and 4, which is 20.

We can either multiply through by 20 or use 20 as a common denominator. 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.

$$ \begin{gather} \small\frac{x+2}{5}\normalsize \gt -\small\frac{3x}{4}\normalsize \\[8pt] \small\frac{4(x+2)}{20}\normalsize \gt -\small\frac{15x}{20}\normalsize \\[8pt] 4(x+2) \gt -15x\\[8pt] 4x+8 \gt -15x \\[8pt] 4x+15x \gt -8\\[6pt] 19x \gt -8\\[6pt] x \gt -\small \frac{8}{19} \normalsize \end{gather} $$

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

This can be done in two 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. You might like to try it the other way, as an exercise.

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

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

This can be done in two 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. You might like to try it the other way, as an exercise.

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

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

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

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

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

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

This can be done in two 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. You might like to try it the other way, as an exercise.

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