Learning Outcomes
- Solve single-step linear inequalities
- Solve multi-step linear inequalities
- Write a solution set in interval notation
- Write solution set in set-builder notation
Key words
- Solution set: a set containing the solutions
Properties of Inequality
A linear inequality is similar to a linear equation in many ways, but do the properties of equality still hold? An inequality like [latex]2\lt 5[/latex] can be visualized as an unbalanced scale:
It seems logical that adding or subtracting the same amount from both sides of the scale will not change the imbalance.
Consider the simple inequality [latex]2 \lt 5[/latex]. Let’s add [latex]4[/latex] to each side and determine if the inequality still holds:
[latex]\begin{equation}\begin{aligned}2 & \lt\ 5 \\ 2 \color{blue}{+4} & \lt 5\color{blue}{+4} \\ 6 & \lt 9 \end{aligned}\end{equation}[/latex]
Adding the same number to both sides keeps the inequality in tact. Let’s try subtracting [latex]3[/latex] from both sides to see what happens:
[latex]\begin{equation}\begin{aligned}2 & \lt\ 5 \\ 2 \color{blue}{-3} & \lt 5\color{blue}{-3} \\ -1 & \lt 2 \end{aligned}\end{equation}[/latex]
Subtracting the same number to both sides keeps the inequality in tact.
The addition and subtraction property of equality applies to inequalities. But what of the multiplication and division properties of equality? Let’s consider the inequality [latex]2\lt 5[/latex] again, and start by multiplying both sides by a positive number.
[latex]\begin{equation}\begin{aligned}2 & \lt\ 5 \\ 2 \color{blue}{\cdot 3} & \lt 5\color{blue}{\cdot 3} \\ 6 & \lt 15 \end{aligned}\end{equation}[/latex]
The inequality still holds. But what about multiplying by a negative number?
[latex]\begin{equation}\begin{aligned}2 & \lt\ 5 \\ 2 \color{blue}{\cdot (-4)} & \lt 5\color{blue}{\cdot (-4)} \\ -8 & \lt -20 \end{aligned}\end{equation}[/latex]
Thuis results in a false statement! To make the statement true, we must flip the [latex]\lt[/latex] sign to a [latex]\gt[/latex] sign because [latex]-8\gt -20[/latex]. The same thing happens if we try to divide by a negative number: we must flip the inequality sign.
Properties of inequality
For all real numbers [latex]a,\,b,\,c[/latex], if [latex]a<b[/latex] then [latex]a+c\lt b+c[/latex].
Adding or subtracting the same term to both sides an inequality keeps the inequality true.
For all real numbers [latex]a,\,b,\,c[/latex] with [latex]c\gt 0[/latex], if [latex]a<b[/latex] then [latex]a\cdot c\lt b\cdot c[/latex].
Multiplying or dividing by a positive term to both sides an inequality keeps the inequality true.
For all real numbers [latex]a,\,b,\,c[/latex] with [latex]c\lt 0[/latex], if [latex]a\lt b[/latex] then [latex]a\cdot c\gt b\cdot c[/latex].
To multiply or divide by a negative term on both sides of an inequality we must reverse the inequality sign.
Solving an linear inequality follows the same rules as solving a linear equation. In the first example, we use the multiplication and division property to isolate the variable.
Example
Solve.
1. [latex]3x\lt 6[/latex] Write your answer in set-builder notation.
2. [latex]-2x - 1\ge 5[/latex] Write your answer in interval notation.
3. [latex]5-x\geq 10[/latex] Graph your answer on a number line.
Solution:
1.
[latex]\begin{equation}\begin{aligned} 3x & < 6 \\ \color{blue}{\frac{1}{3}}\left( 3x \right) & < \left ( 6\right )\color{blue}{\frac{1}{3}}\;\;\;\;\;\;\;\;\;\;\text{Multiply both sides by }\frac{1}{3} \text{ (or divide both sides by }3) \\ x & \lt 2 \end{aligned}\end{equation}[/latex]
Answer: [latex]\{\;x\;\large |\normalsize\;\;x \lt 2,\;x\in\mathbb{R}\;\}[/latex]
2.
[latex]\begin{equation}\begin{aligned}-2x - 1 & \ge 5\;\;\;\;\;\;\;\;\;\;\text{Add 1 to both sides} \\ -2x & \ge 6\\ \color{blue}{\left(-\frac{1}{2}\right)}\left(-2x\right) & \color{blue}{\le} \left(6\right)\color{blue}{\left(-\frac{1}{2}\right)}\;\;\;\;\;\;\;\;\;\;\text{Multiply both sides by }-\frac{1}{2} \text{ and reverse the inequality sign} \\ x & \le -3\end{aligned}\end{equation}[/latex]
Answer: [latex]x\in \left ( -\infty,\,-3\right ][/latex]
3.
[latex]\begin{equation}\begin{aligned}5-x & \geq10 \\ -x & \geq 5\\ \color{blue}{\left(-1\right)}\left(-x\right) & \color{blue}{\leq} \left(5\right)\color{blue}{\left(-1\right)}\;\;\;\;\;\;\;\;\;\;\text{Multiply both sides by } -1\text{ and reverse the sign}\\ x & \leq-5\end{aligned}\end{equation}[/latex]
Answer: 
Try It
Solve:
1. [latex]4x\ge -8[/latex] Write your answer in set-builder notation.
2. [latex]-5x\gt 4[/latex] Write your answer in interval notation.
3. [latex]-3x\lt -9[/latex] Graph your answer on a number line.
In our next example, we will use the addition property to solve inequalities.
Example
Solve. Write the answer in interval notation.
1. [latex]x - 15\lt 4[/latex]
2. [latex]6\ge x - 1[/latex]
3. [latex]x+7\gt 9[/latex]
Solution:
1.
[latex]\begin{equation}\begin{aligned} x - 15 & \lt 4 \\ x - 15 \color{blue}{+15} & \lt 4 \color{blue}{+15}\;\;\;\;\;\;\;\;\;\; \text{Add 15 to both sides.}\\ x & \lt 19 \end{aligned}\end{equation}[/latex]
Answer: [latex]x\in\left ( -\infty,\,19\right )[/latex]
2.
[latex]\begin{equation}\begin{aligned}6 & \geq x - 1 \\ 6\color{blue}{+1} & \geq x - 1 \color{blue}{+1} \;\;\;\;\;\;\;\;\;\; \text{Add 1 to both sides}. \\ 7 & \geq x \end{aligned}\end{equation}[/latex]
Answer: [latex]x\in\left ( -\infty,\,7\right ][/latex]
3.
[latex]\begin{equation}\begin{aligned}x+7 & \gt 9\\ x+7 \color{blue}{- 7} & \gt 9\color{blue}{- 7}\;\;\;\;\;\;\;\;\;\; \text{Subtract 7 from both sides}.\\x & \gt 2\end{aligned}\end{equation}[/latex]
Answer: [latex]x\in\left [ 2,\,\infty\right )[/latex]
Try It
Solve. Write the answer in interval notation.
1. [latex]x-8\lt -6[/latex]
2. [latex]x+7\gt 7[/latex]
3. [latex]x+\frac{2}{5}\geq -\frac{2}{5}[/latex]
The following video shows examples of solving single-step inequalities using the multiplication and addition properties.
The following video shows examples of solving inequalities with the variable on the right side.
Solving Multi-Step Inequalities
As the previous examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations. We just have to remember to reverse the sign if we multiply or divide by a negative term. To isolate the variable and solve, we combine like terms and perform operations with the multiplication and addition properties.
Example
Solve the inequality: [latex]13 - 7x\ge 10x - 4[/latex]. Write the solution in interval notation.
Solution
Solving this inequality is similar to solving an equation up until the last step.
The solution set is given by the interval [latex]\left(-\infty ,1\right][/latex].
Try It
In the next example, we solve an inequality that contains fractions. Notice how we need to reverse the inequality sign at the end because we multiply by a negative.
Example
Solve the following inequality and write the answer in interval notation: [latex]-\frac{3}{4} x\ge -\frac{5}{8} +\frac{2}{3} x[/latex].
Solution
Instead of working with the fractions in this last example, we could clear the fractions by multiplying both sides by a common denominator.
Example
Solve the following inequality and write the answer in interval notation: [latex]-\frac{3}{4} x\ge -\frac{5}{8} +\frac{2}{3} x[/latex].
Solution
The least common multiple of 4, 8 and 3 is 24. We will multiply both sides of the inequality by 24. This requires applying the distributive property to the two terms on the right side of the inequality.
Try It
Solve the following inequality and write the answer in interval notation: [latex]\frac{1}{3} x\le -\frac{4}{5} +\frac{3}{4} x[/latex].
