Algebra 1/Unit 7: Inequalities

Algebra 1/Unit 7: Inequalities

Welcome to Week 7 of the Algebra 1 course! This week, we will cover the foundations and basics of inequalities, which are mathematical statements that describe one quantity being larger or smaller than another quantity, rather than being exactly equal. Inequalities are important in general because they allow us to represent ranges of possible values (for example: you can describe all numbers that are greater than 5, or all scores needed to pass a test). Many real-world scenarios like budgets, speed limits, and minimum/maximum requirements for anything (like computer specifications to play the latest game at 4K 120fps+), among many other examples, are naturally expressed using inequalities.

More specifically, this unit will cover:

• Understanding the inequality symbols, $>,<,$ $\geq$, $\leq$ and what they mean
• Solving basic linear inequalities in one variable
• Graphing inequality solutions on a number line
• Writing/Solving compound inequalities (using "and" / "or")
• An introduction to linear inequalities in two variables and how to graph them on the coordinate plane

Introduction

Let's first begin with a simple refresher on how to solve linear equations as they tie into inequalities. Say we have an equation like $x+2=7$, to solve the equation, we find specific value(s) of $x$ to make the equation/statement true (in this case $x=5$). An inequality, on the other hand, expresses or conveys a relation where one side can be greater than or less than the other side. For example, $x+2>7$ means that we are looking for all values of $x$ on the left-hand side that make $x+2$ greater than ($>$) the right hand side value of 7. In this case, the statement would be true for any $x>5$ ($x$ value greater than 5). We can see that instead of getting a single solution like how we would get when solving for linear equations, inequalities usually have infinitely many solutions, meaning the solution is a range or interval of values.

Inequalities are written using special symbols to show comparisons between two sides. Let's first dive into understanding what each inequality symbols means.

Understanding Inequality Symbols ($>$, $<$, $\geq$, $\leq$)

The four main inequality symbols and their meanings are (when reading an inequality from left to right):

• > (greater than) – indicates that the left side is larger than the right side. Example: 7 > 5 means 7 is greater than 5 (which is a true statement). If we write x > 3, it means x must be a value greater than 3 (so x could be 4, 5, 10, 3.1, etc., but not 3 itself). This is called a strict inequality (x cannot equal 3).
• < (less than) – indicates that the left side is smaller than the right side. Example: 2 < 9 means 2 is less than 9. Similarly, y < 0 means y must be less than 0 (y could be -1, -5.5, etc., but not 0). This is also a strict inequality.
• ≥ (greater than or equal to) – indicates that the left side is larger than or equal to the right side. Example: x ≥ 4 means x is greater than or equal to 4 or x is at least 4. So x could be 4 or any number greater (5, 6, 4.2, etc.). Here the underscore/underline below the symbol means equality is allowed. Hence, 4 is a valid solution since 4 ≥ 4 (four is indeed greater than or equal to four).
• ≤ (less than or equal to) – indicates that the left side is smaller than or equal to the right side. Example: y ≤ 0 means y is less than or equal to 0 or y is at most 0 . So y could be 0 or any number less than 0 (−1, −2.5, etc.). Equality is also allowed here, meaning y = 0 is a valid solution.

Note: The symbols > and < represent strict inequalities (meaning the two sides are not equal), whereas ≥ and ≤ are non-strict (inclusive) inequalities that allow equality as a possibility. In everyday language, $\geq$ is often phrased as “at least” and $\leq$ as “at most.”

Solving Linear Inequalities in One Variable

Solving an inequality in one variable (like $x$) is very similar to solving a linear equation as we have mentioned prior. The goal is essentially the same, you want to isolate the variable on one side by using algebraic operations. Then you want to either add, subtract, multiply, or divide both sides of an inequality by the same number just like you would do in a linear equation. The solution will often be another inequality that describes the range of values that will make the original inequality/statement true.

Key Rule: If you multiply or divide both sides of an inequality by a negative number (-1, -4, -5, etc.), you must flip the inequality symbol. So in other words, $>$ becomes $<$, $<$ becomes $>$, $\geq$ becomes $\leq$, and $\leq$ becomes $\geq$. This is because multiplying or dividing by a negative reverses the order (Adding or subtracting a number does not flip the sign)

Steps to solve a linear inequality:

• Simplify each side of the inequality (expand and combine like terms if needed)
• Use addition or subtraction to move/cancel terms and get the variable term by itself on one side of the inequality
• Use multiplication or division to solve for the variable (Remember to flip the inequality symbol if you multiply/divide by a negative value)

After solving the inequality, you can describe the solution set in words. For example, if a linear inequality has a solution set of $x>2$ you can say the solution is "all x values greater than 2", or you could simply keep it in inequality form if it is easier for you.

Example 1: Solve the inequality $3x+9>30$.

1. Subtract 9 from both sides of the inequality symbol to isolate the term with x:

$3x+9>30$ $3x+9-9>30-9$ $3x>21$

1. Divide both sides by 3 to solve for x. (since we are not dividing or multiply by a negative here, you do NOT have to flip the inequality symbol): $3x\div (3)>21\div (3)$ $x>7$

Hence, the solution is $x>7$. This means that any number that is greater than 7 will satisfy the original inequality and make it a true statement. We can also test this with a quick check: if we set $x=8$ (which is greater than 7), then plug that into left side of the inequality where the the $x$ variable is, we will get: $3(8)+9=24+9=33$, and so the left hand side would be 33, resulting in the inequality reading: $33>30$ which is a true statement. Now what would happen if we did the opposite? Let's set $x$ to be less than 7, let's choose $x=6$. Then when plugging that in, we get $3(6)+9=18+9=27$ on the lefthand side. This makes the inequality read $27>30$ which is not a true statement (as 27 is not greater in value than 30). Also, if you were to plug in $x=7$ the inequality would read $30>30$ after multiplying and combining like terms. This statement is also false because 30 is not greater than 30, it is equal to 30. If the inequality symbol was $\geq$ or $\leq$ then the inequality statement would be true (since those statements say "greater than or equal to" and "less than or equal to") This emphasizes and shows how only $x$ values strictly greater than 7 make the given inequality statement true.

Example 2: Solve the inequality $-2x+5>15$.

1. Subtract 5 from both sides of the inequality symbol to isolate the term with x: $-2x+5>15$ $-2x+5-5>15-5$ $-2x>10$
2. Divide both sides by $-2$ to solve for x. And importantly, since we are dividing by a negative in this case, you must flip the inequality sign/symbol (after the division step): $-2x\div (-2)>10\div (-2)$ $x<-5$ (we flipped the sign here)

As we can see the solution is $x<-5$. This means that any $x$ value less than -5 will satisfy/make the original inequality statement $-2x+5>15$ true. Just like the last example, you can verify this by quickly plugging in a number that is less than -5. Let's choose $x=-6$ (remember this is smaller than -5 because its further away from 0. Try to imagine you are going 6 spaces to the left of 0 on a number line if you are having trouble). Then plugging in $x=-6$ to the inequality will make the lefthand side equal to: $-2(-6)+5=12+5=17$ and the inequality statement would be $17>15$, which is true. Feel free to plug in other values less than -5 to confirm that the inequality is true and also try to look at why values that are greater than or equal to 5 make the inequality false (like we have already exemplified in Example 1). There you go, you now know the basics of solving for inequalities! If you are still not comfortable solving these sorts of problems, don't worry, with practice you will get the hang of it in no time.

Practice Problems (Linear Inequalities)

Below are some problems for you to practice on: Solve the Following Inequalities

1

$3x>15$

Solution: $x$ $>$

2

$2x+7<21$

Solution: $x$ $<$

3

$-5x+10>15$

Solution: $x$ $<$

4

$8x-9>15$

Solution: $x$ $>$

5

$-7x+13<62$

Solution: $x$ $>$

Writing and Solving Compound Inequalities

Now that we have gone over how to identify and solve linear inequalities, we can go a step further and delve into compound inequalities. Don't worry, this sounds hard, but it's very simple once you get the hang of it.

So what exactly is a compound inequality? A compound inequality essentially combines two inequalities joined by the words "and" or "or". This allows us to describe ranges of values bounded either by two specific values or to describe solutions that satisfy one condition or another.

More specifically, these are what each words mean in the context of inequalities:

• "AND" (formally called Conjunction in mathematics): The solution must satisfy both conditions (both inequalities in this case) at the same time. This usually describes a "between" situation. To give you a clearer idea, take the example of "x is greater than 2 and less than 5". This example is simply saying that the value of $x$ can be between 2 and 5, of course excluding 2 and 5 themselves as we know that greater than and less than are both strict cases (for more clarification look at the Understanding Inequality Symbols section at the beginning of the page). We can write this as $2<x<5$ (a combined inequality) or as two separate inequalities with the "and" keyword in between them: $x>2$ and $x<5$. Note that for the first option ($2<x<5$) we are reading the inequality based on the position of $x$, meaning that we read it as "x is greater than 2 and less than 5" and not "2 is less than x and x is less than 5". Technically the second way of saying it is not wrong, but it is easier and a convention to say it the first way. Moving on, the solution set for an "and" compound inequality is the intersection of the solution sets of each individual inequality (i.e., only the values that make both inequality statements true).

• "OR" (formally called Disjunction in mathematics): The solution can satisfy either one condition or the other (or both), meaning either both inequality statements can be true, or one of the inequalities can be satisfied as true while the other is false, and vice versa. This keyword is usually used to describe two separate regions of value. For example, say we have the compound inequality statement "y is less than 0 or greater than 9". This means $y$ can take on values either below 0 or above 9. We can write this as $y<0$ or $y>9$. The solution set for an "or" inequality is the union of each individual solution set (any value that works/satisfies for at least one of the inequalities).

If you're still confused, don't worry, we will do some problems together to make it more clear in the "Solving Compound Inequalities" section below:

Solving Compound Inequalities:

• For an "and" inequality in the combined form (e.g. $4<3x+1\leq 13$), you can either choose to solve it in one go by performing your standard algebra operations (as you would in solving a linear equation) on all three parts of the compound inequality (lefthand side, middle, and righthand side) or if you prefer, you can split the compound inequality into two separate inequalities, solve for each of them, and then find the overlapping values in the two solution sets.
• For an "or" inequality, you have to solve each inequality separately, then combine the solutions by sticking the "or" keyword between them (this is not looking for overlapping values, you are just taking all values from both, which is called union)

NOTE: Be extremely careful about flipping signs in compound inequalities as well. They essentially share the same rules as singular linear inequalities, just with one or more extra inequalities combined into one. So remember in this case, if you multiply or divide all parts of an "and" compound inequality by a negative, then each of the inequality signs/symbols flip.

Example of "AND" (Conjunction) inequality: Solve the compound inequality $4<3x+1\leq 13$:

1. Before even solving the inequality, try and split the inequality into two separate ones to clearly see what the conditions are. In this case, the inequality means $4<3x+1$ and $3x+1\leq 13$ simultaneously. Now, we will solve this inequality by handling all parts at the same time (isolating the variable term), step by step. First, subtract 1 from all parts of the inequality: $4-1<3x+1-1\leq 13-1$ $3<3x\leq 12$
2. Next, divide all parts by 3 (since it is a positive number we do not have to flip the signs of the inequalities): $3\div 3<3x\div 3\leq 12\div 3$ $1<x\leq 4$

Now that we have isolated the variable ($x$ in this case) term by itself and simplified all terms, we are done! The solution can be written as $1<x\leq 4$, which means that "x is greater than 1 and less than or equal to 4 (similarly you could say up to 4 instead of less than or equal to as they are the same thing). In other words, $x$ is between 1 and 5, not including 1 but including 5. If you wanted to, you could also express the solution as "$x>1$ and $x\leq 4$", but it's easier and quicker to leave the solution as what we had earlier.

Example of "OR" (Disjunction) inequality: Solve the compound inequality $x-2\leq -5$ or $3x>21$:

Here we have two separate inequalities connected by "or". You simply have to solve each one individually then combine them at the end with the "or" keyword:

A) Solve $x-2\leq -5$

1. Add 2 to both sides $x-2+2\leq -5+2$ $x\leq -3$ (Done with the first inequality)

B) Solve $3x>21$

1. Divide both sides by 3 $3x\div 3>21\div 3$ $x>7$

Now that we have gotten the individual solutions for each of the inequalities, we can simply say the solution is $x\leq -3$ or $x>7$. This means that any $x$ value that is at most (less than or equal to) $-3$ or greater than $7$ will satisfy at least one of the inequalities. Sidenote, numbers that are less than or equal to -3 satisfy the first part/inequality and numbers greater than 7 satisfy the second inequality. This means that if a given input value satisfies at least one of the inequalities/parts, then that input is a valid solution for the entire statement ($x-2\leq -5$ or $3x>21$ in this case). For instance, if you say a solution is $x=10$, that is valid/accepted because it makes the the second inequality true. Similarly, saying $x=-5$ also is valid because it makes the first inequality true. The only thing is, you cannot have a value that does not satisfy any of the inequalities/parts. For example, you cannot say a solution to the compound inequality is $x=0$ because plugging $0$ into the first inequality yields $-2\leq -5$ which is not true and plugging $0$ into the second inequality results in $0>21$ which is also not true. Since neither of the two inequalities are satisfied with $x=0$, it is not a solution to the compound inequality.

There you have it, now you know how to solve different kinds of compound inequalities. Feel free to practice more if you do not understand or feel confident enough in solving them. More practice problems are down below:

Practice Problems (Compound Inequalities)

1

$6<2x+4<16$

Solution: $<x$ $<$

2

$5x-3<7$ or $7x-3>32$

Solution: $x$ $<$ or $x>$

3

What is the solution to $5>-3x+2\geq -7$

	$-1>x\geq 3$
	$-1<x\leq 3$
	$-1<x<3$
	$-1<x\geq 3$

4

What is the solution to $-2x+2\leq -20$ or $-3x-4\geq 23$

	$x\geq 11$ or $x\leq -9$
	$x\geq -11$ or $-9>x$
	$x\leq 11$ or $x\geq -9$
	$x>11$ or $x<-9$

5

(BONUS) What is the solution to ($-2<x+1\leq 3$) or ($2x-5>1$ and $x+4\leq 8$)

	$-3>x\geq 2$ or $3>x\geq 4$
	$-3<x<2$ and $3\leq x\leq 4$
	$-3\leq x\leq 2$ or $3<x<4$
	$-3<x\leq 2$ or $3<x\leq 4$

6

(BONUS 2 Challenge) What is the solution to ($4-x\geq 2$) and ($2x-5<1$ or $-3x+4\geq 1$)

	$x<3$ and $-3x+4\geq 1$
	$x\leq 1$ or $x<3$
	$x\leq 2$
	$x>2$

Linear Inequalities in Two Variables (Brief Introduction)

Just like linear equations that have two variables (i.e. $y=5x+2$) represent a line on the coordinate plane, a linear inequality in two variables (i.e. $y>5x+2$) represents a region of the coordinate plane. The solution to a two-variable inequality is not a single number like in a linear equation, but rather, it is all the ordered pairs $(x,y)$ that make the inequality true.

To learn more about linear inequalities in two variables and how to specifcally graph them, along with learning how to graph regular one-variable linear inequalities and compound inequalities, check out Unit 8: Graphing Inequalities.

References

• Khan Academy “Solving Inequalities.” https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq (Guide for solving and interpreting one-variable linear inequalities.)
• OpenStax Stewart, J., et al. Intermediate Algebra (OpenStax, 2016). https://openstax.org/details/books/intermediate-algebra (Helping with definitions, examples, and exercises on inequality notation and solution sets.)

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