In other lessons, we learned how to solve one-step, two-step, and multi-step equations. In solving linear inequalities, the approach is very similar. Instead of using the equality symbol \( = \), we use inequality symbols like greater than \( > \), and less than \(<\). We just need to be extra careful with inequality symbols because they behave a bit differently. Just remember that the direction of the inequality symbol **switches direction** when we multiply or divide both sides of the inequality by a **negative** number. That's the nuance we have to consider at all times.

Example 1: Solve the inequality below.

$$5x - 2 > 3x + 6$$

Treat this just like a regular equation. We solve the inequality by isolating the variable on one side of the inequality symbol. For uniformity, let's agree to keep the variables to the left while the numbers to the right side of the inequality.

The first thing we can do is to eliminate the \(-2\) on the left side by adding \(2\) to both sides of the inequality. Then remove the \(3x\) on the right by subtracting both sides by \(3x\). Finally, divide both sides by the coefficient of the variable which is \(2\). Since we are dividing both sides by a positive number, the orientation of the inequality remains the same.

$$\eqalign{

5x - 2 + 2 &> 3x + 6 + 2 \cr

5x &> 3x + 8 \cr

5x - 3x& > 3x - 3x + 8 \cr

2x& > 8 \cr

{{2x} \over 2}& > {8 \over 2} \cr

x& > 4 \cr} $$

We can also write the solution in terms of interval notation.

$$\left( {4,\infty } \right)$$

Example 2: Solve the inequality below.

$$x - 3x + 11 \ge 4x - 19$$

Notice that there are two terms on the left side that have x's. Let's combine them first to simplify the problem. We have \(x - 3x = - 2x\).

$$ - 2x + 11 \ge 4x - 19$$

We can subtract both sides of the inequality by \(4x\) to move the variables to the left. Then, subtract both sides by \(11\) to move the numbers to the right of the inequality symbol.

$$\eqalign{

- 2x + 11 &\ge 4x - 19 \cr

- 2x - 4x + 11 &\ge 4x - 4x - 19 \cr

- 6x + 11 &\ge - 19 \cr

- 6x + 11 - 11 &\ge - 19 - 11 \cr

- 6x &\ge - 30 \cr} $$

Finally, divide both sides by the coefficient of the variable which is \(-6\). Since we are dividing both sides by a negative number, the direction of the inequality symbol switches from left to right.

$$\eqalign{

- 6x &\ge - 30 \cr

{{ - 6x} \over { - 6}} &\color{red}{\le} {{ - 30} \over { - 6}} \cr

x &\color{red}{\le} 5 \cr} $$

This is the solution expressed in interval notation.

$$\left( { - \infty ,5} \right]$$

Example 3: Solve the inequality below.

$$7 - 6\left( {x + 4} \right) < 13 + 3\left( {1 - x} \right)$$

Note that there is a parenthesis on each side of the inequality. It's obvious that we will apply the Distributive Property of Multiplication over Addition here to get rid of them. Then, we combine like terms.

$$\eqalign{

7 - 6\left( {x + 4} \right) &< 13 + 3\left( {1 - x} \right) \cr

7 - 6x - 24 &< 13 + 3 - 3x \cr

- 6x - 17 &< 16 - 3x \cr} $$

From this point, let's solve the inequality as usual. Add both sides by \(17\). Then, to get rid of the variable on the right side, we add \(3x\) to both sides. Finally, divide both sides by \(-3\). Since we are dividing by a negative number, make sure to flip the direction of the inequality symbol from less than to greater than.

$$\eqalign{

- 6x - 17 + 17 &< 16 + 17 - 3x \cr

- 6x &< 33 - 3x \cr

- 6x + 3x &< 33 - 3x + 3x \cr

- 3x &< 33 \cr

{{ - 3x} \over { - 3}} &\color{red}{>} {{33} \over { - 3}} \cr

x &\color{red}{> } - 11 \cr} $$

Here is the solution in interval notation.

$$\left( { - 11,\infty } \right)$$