Solving Linear Inequalities

In previous lessons, we explored how to solve one-step, two-step, and multi-step equations. The process for solving linear inequalities is quite similar, with the key difference being the use of inequality symbols, such as greater than \(>\) and less than \(<\), instead of the equality symbol \(=\). However, it’s important to be cautious with these symbols because they behave differently in certain situations. Specifically, the direction of the inequality symbol reverses when we multiply or divide both sides of the inequality by a negative number. This is a crucial detail to keep in mind.

Example 1: Solve the inequality below.

$$5x – 2 > 3x + 6$$

We treat this like a standard equation, focusing on isolating the variable on one side of the inequality. To maintain consistency, let’s keep the variables on the left side and the numbers on the right side of the inequality.

First, eliminate the \(-2\) on the left side by adding \(2\) to both sides. Next, remove the \(3x\) on the right side by subtracting \(3x\) from both sides of the inequality. Finally, divide both sides by the variable’s coefficient, which is \(2\). Since we’re dividing by a positive number, the direction of the inequality remains unchanged.

$$\eqalign{
   5x – 2 + 2 &> 3x + 6 + 2  \cr 
   5x &> 3x + 8  \cr 
   5x – 3x& > 3x – 3x + 8  \cr 
   2x& > 8  \cr 
   {{2x} \over 2}& > {8 \over 2}  \cr 
   x& > 4 \cr} $$

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

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

Example 2: Solve the inequality below.

$$x – 3x + 11 \ge 4x – 19$$

Observe that there are two terms on the left side that contain x. Let’s start by combining them to simplify the equation. When we do this, we get \(x – 3x = -2x\).

$$ – 2x + 11 \ge 4x – 19$$

We can subtract \(4x\) from both sides of the inequality to bring the variable terms to the left side. Next, subtract \(11\) from both sides to move the constant terms to the right side of the inequality.

$$\eqalign{
    – 2x + 11 &\ge 4x – 19  \cr 
    – 2x – 4x + 11 &\ge 4x – 4x – 19  \cr 
    – 6x + 11 &\ge  – 19  \cr 
    – 6x + 11 – 11 &\ge  – 19 – 11  \cr 
   – 6x &\ge  – 30 \cr} $$

Finally, divide both sides by the coefficient of the variable, which is \(-6\). Because we are dividing by a negative number, the direction of the inequality symbol reverses.

$$\eqalign{
    – 6x &\ge  – 30  \cr 
   {{ – 6x} \over { – 6}} &\color{red}{\le} {{ – 30} \over { – 6}}  \cr 
   x &\color{red}{\le} 5 \cr} $$

This is the solution expressed in interval notation.

$$\left( { – \infty ,5} \right]$$

Example 3: Solve the inequality below.

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

Notice that there are parentheses on both sides of the inequality. Clearly, we will use the Distributive Property of Multiplication over Addition to eliminate them. After that, we can combine like terms.

$$\eqalign{
   7 – 6\left( {x + 4} \right) &< 13 + 3\left( {1 – x} \right)  \cr 
   7 – 6x – 24 &< 13 + 3 – 3x  \cr 
    – 6x – 17 &< 16 – 3x \cr} $$

Let’s proceed with solving the inequality step by step. First, add \(17\) to both sides. Next, to eliminate the variable on the right side, add \(3x\) to both sides. Finally, divide both sides by \(-3\). Remember, when dividing by a negative number, you must reverse 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.

$$\left( { – 11,\infty } \right)$$