We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > College Algebra: Co-requisite Course

Linear Inequalities in One Variable

Learning Objectives

  • Use the addition and multiplication properties to solve algebraic inequalities
  • Express solutions to inequalities graphically, with interval notation, and as an inequality
  • Simplify and solve algebraic inequalities using the distributive property to clear parentheses and fractions

Using the Properties of Inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol. There are three ways to represent solutions to inequalities: an interval, a graph, and an inequality. Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality. When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities.

A General Note: Properties of Inequalities

[latex]\begin{array}{ll}\text{Addition Property}\hfill& \text{If }a< b,\text{ then }a+c< b+c.\hfill \\ \hfill & \hfill \\ \text{Multiplication Property}\hfill & \text{If }a< b\text{ and }c> 0,\text{ then }ac< bc.\hfill \\ \hfill & \text{If }a< b\text{ and }c< 0,\text{ then }ac> bc.\hfill \end{array}[/latex]

These properties also apply to [latex]a\le b[/latex], [latex]a>b[/latex], and [latex]a\ge b[/latex].
The following table illustrates how the multiplication property is applied to inequalities, and how multiplication by a negative reverses the inequality:
Start With Multiply By Final Inequality
[latex]a>b[/latex] [latex]c[/latex] [latex]ac>bc[/latex]
[latex]5>3[/latex] [latex]3[/latex] [latex]15>9[/latex]
[latex]a>b[/latex] [latex]-c[/latex] [latex]-ac<-bc[/latex]
[latex]5>3[/latex] [latex]-3[/latex] [latex]-15<-9[/latex]
The following table illustrates how the division property is applied to inequalities, and how dividing by a negative reverses the inequality:
Start With Divide By Final Inequality
[latex]a>b[/latex] [latex]c[/latex] [latex] \displaystyle \frac{a}{c}>\frac{b}{c}[/latex]
[latex]4>2[/latex] [latex]2[/latex] [latex] \displaystyle \frac{4}{2}>\frac{2}{2}[/latex]
[latex]a>b[/latex] [latex]-c[/latex] [latex] \displaystyle -\frac{a}{c}<-\frac{b}{c}[/latex]
[latex]4>2[/latex] [latex]-2[/latex] [latex] \displaystyle -\frac{4}{2}<-\frac{2}{2}[/latex]
In the first example, we will show how to apply the multiplication and division properties of equality to solve some inequalities.

Example: Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:
  1. [latex]x - 15<4[/latex]
  2. [latex]6\ge x - 1[/latex]
  3. [latex]x+7>9[/latex]

Answer: The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

  1. [latex]\begin{array}{ll}x - 15<4\hfill & \hfill \\ x - 15+15<4+15 \hfill & \text{Add 15 to both sides.}\hfill \\ x<19\hfill & \hfill \end{array}[/latex]
  2. [latex]\begin{array}{ll}6\ge x - 1\hfill & \hfill \\ 6+1\ge x - 1+1\hfill & \text{Add 1 to both sides}.\hfill \\ 7\ge x\hfill & \hfill \end{array}[/latex]
  3. [latex]\begin{array}{ll}x+7>9\hfill & \hfill \\ x+7 - 7>9 - 7\hfill & \text{Subtract 7 from both sides}.\hfill \\ x>2\hfill & \hfill \end{array}[/latex]

Try It

Solve [latex]3x - 2<1[/latex].

Answer: [latex-display]x<1[/latex-display]

Example: Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:
  1. [latex]3x<6[/latex]
  2. [latex]-2x - 1\ge 5[/latex]
  3. [latex]5-x>10[/latex]

Answer:

  1. [latex]\begin{array}{l}3x<6\hfill \\ \frac{1}{3}\left(3x\right)<\left(6\right)\frac{1}{3}\hfill \\ x<2\hfill \end{array}[/latex]
  2. [latex]\begin{array}{ll}-2x - 1\ge 5\hfill & \hfill \\ -2x\ge 6\hfill & \hfill \\ \left(-\frac{1}{2}\right)\left(-2x\right)\ge \left(6\right)\left(-\frac{1}{2}\right)\hfill & \text{Multiply by }-\frac{1}{2}.\hfill \\ x\le -3\hfill & \text{Reverse the inequality}.\hfill \end{array}[/latex]
  3. [latex]\begin{array}{ll}5-x>10\hfill & \hfill \\ -x>5\hfill & \hfill \\ \left(-1\right)\left(-x\right)>\left(5\right)\left(-1\right)\hfill & \text{Multiply by }-1.\hfill \\ x<-5\hfill & \text{Reverse the inequality}.\hfill \end{array}[/latex]

Try It

Solve [latex]4x+7\ge 2x - 3[/latex].

Answer: [latex]x\ge -5[/latex]

Solving Inequalities in One Variable Algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

Example: Solving an Inequality Algebraically

Solve the inequality: [latex]13 - 7x\ge 10x - 4[/latex].

Answer: Solving this inequality is similar to solving an equation up until the last step.

[latex]\begin{array}{ll}13 - 7x\ge 10x - 4\hfill & \hfill \\ 13 - 17x\ge -4\hfill & \text{Move variable terms to one side of the inequality}.\hfill \\ -17x\ge -17\hfill & \text{Isolate the variable term}.\hfill \\ x\le 1\hfill & \text{Dividing both sides by }-17\text{ reverses the inequality}.\hfill \end{array}[/latex]
The solution set is given by the interval [latex]\left(-\infty ,1\right][/latex], or all real numbers less than and including 1.

Try It

Solve the inequality and write the answer using interval notation: [latex]-x+4<\frac{1}{2}x+1[/latex].

Answer: [latex]\left(2,\infty \right)[/latex]

Example: Solving an Inequality with Fractions

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].

Answer: We begin solving in the same way we do when solving an equation.

[latex]\begin{array}{ll}-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x\hfill & \hfill \\ -\frac{3}{4}x-\frac{2}{3}x\ge -\frac{5}{8}\hfill & \text{Put variable terms on one side}.\hfill \\ -\frac{9}{12}x-\frac{8}{12}x\ge -\frac{5}{8}\hfill & \text{Write fractions with common denominator}.\hfill \\ -\frac{17}{12}x\ge -\frac{5}{8}\hfill & \hfill \\ x\le -\frac{5}{8}\left(-\frac{12}{17}\right)\hfill & \text{Multiplying by a negative number reverses the inequality}.\hfill \\ x\le \frac{15}{34}\hfill & \hfill \end{array}[/latex]
The solution set is the interval [latex]\left(-\infty ,\frac{15}{34}\right][/latex].

Try It

Solve the inequality and write the answer in interval notation: [latex]-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x[/latex].

Answer: [latex-display]\left[-\frac{3}{14},\infty \right)[/latex-display]

Simplify and solve algebraic inequalities using the distributive property

As with equations, the distributive property can be applied to simplify expressions that are part of an inequality. Once the parentheses have been cleared, solving the inequality will be straightforward.

Example

Solve for x. [latex]2\left(3x–5\right)\leq 4x+6[/latex]

Answer: Distribute to clear the parentheses.

[latex] \displaystyle \begin{array}{r}\,2(3x-5)\leq 4x+6\\\,\,\,\,6x-10\leq 4x+6\end{array}[/latex]

Subtract 4x from both sides to get the variable term on one side only.

[latex]\begin{array}{r}6x-10\le 4x+6\\\underline{-4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4x}\,\,\,\,\,\,\,\,\,\\\,\,\,2x-10\,\,\leq \,\,\,\,\,\,\,\,\,\,\,\,6\end{array}[/latex]

Add 10 to both sides to isolate the variable.

[latex]\begin{array}{r}\\\,\,\,2x-10\,\,\le \,\,\,\,\,\,\,\,6\,\,\,\\\underline{\,\,\,\,\,\,+10\,\,\,\,\,\,\,\,\,+10}\\\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\le \,\,\,\,\,16\,\,\,\end{array}[/latex]

Divide both sides by 2 to express the variable with a coefficient of 1.

[latex]\begin{array}{r}\underline{2x}\le \,\,\,\underline{16}\\\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\le \,\,\,\,\,8\end{array}[/latex]

Answer

Inequality: [latex]x\le8[/latex] Interval: [latex]\left(-\infty,8\right][/latex] Graph: The graph of this solution set includes 8 and everything left of 8 on the number line. Number line with the interval (-oo,8] graphed

Check the solution.

Answer: First, check the end point 8 in the related equation.

[latex] \displaystyle \begin{array}{r}2(3x-5)=4x+6\,\,\,\,\,\,\\2(3\,\cdot \,8-5)=4\,\cdot \,8+6\\\,\,\,\,\,\,\,\,\,\,\,2(24-5)=32+6\,\,\,\,\,\,\\2(19)=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\38=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Then, choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 0.

[latex] \displaystyle \begin{array}{l}2(3\,\cdot \,0-5)\le 4\,\cdot \,0+6?\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2(-5)\le 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-10\le 6\,\,\end{array}[/latex]

[latex-display]x\le8[/latex] is the solution to [latex]\left(-\infty,8\right][/latex-display]

In the following video, you are given an example of how to solve a multi-step inequality that requires using the distributive property. https://youtu.be/vjZ3rQFVkh8

Try It

[ohm_question]143594[/ohm_question]
 

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program.. Provided by: Monterey Institute of Technology and Education. Authored by: . Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • College Algebra. Provided by: Lumen Learning Authored by: Jay Abramson, et al... License: CC BY: Attribution.
  • Question ID# 92604, 92605, 92606, 92607, 92608, 92609.. Authored by: Michael Jenck. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Question ID# 72891. Authored by: Alyson Day. License: CC BY: Attribution.
  • Solve a Linear Inequality Requiring Multiple Steps (One Var). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.