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# 2.5: Solving Linear Inequalities in One Variable

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## The Rules Used to Solve Linear Inequalities in One Variable

Objectives

The lesson objectives for Solving Linear Inequalities in One Variable are:

• The rules used to solve linear inequalities
• Solving linear inequalities algebraically
• Graphing the solution set of linear inequalities on a number line

Introduction

In the previous lessons you rules to solve for algebraic expressions. The rules were that you should always:

1. Adding the same number to both sides
2. Subtract the same number from both sides
3. Multiply each side by the same number
4. Divide each side by the same number
5. Remove brackets first using the distributive property (if brackets exist)
6. Get a common denominator if there are fractions

In this lesson, you are going to learn about linear inequations. Linear inequations have a different form than linear equations. You learned that linear equations had the general form of $ax + b = c$, where $a \ne 0$. Linear inequations can have one of four forms: $ax + b > c, ax + b < c, ax + b \ge c,$ or $ax + b \le c$. Notice the difference? That’s right! There is no equal sign. The equal sign is replaced with an inequality sign.

When you solve for a linear inequality, you follow the same rules as you would for a linear inequality, however, you must remember one big rule. If you divide or multiply by a negative number (look at Numbers 3, and 4 above), the sign of the inequality is reversed.

Watch This

Guidance

Janet holds up a card that reads $2x + 6 = 16$. Donna holds up a card that reads $2x + 6 > 16$. Andrew says they are not the same but Donna argues with him. Show, using an example, that Andrew is correct.

Andrew could use a real world example. For example, say Andrew help out two $5 bills and six$1 bills. Andrew holds Janet’s card and says is this true?

Now let’s try it with Donnas’ inequality.

This amount of money is not greater than $16; it is just equal to$16. Andrew then proved the two are not equal.

Example A

In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 8 in for $p$?

Equation Inequality Is the inequality still true?
$2p + 4 = 20$ $2p + 4 < 20$ ?
$2p + 4 - 4 = 20 - 4$
$2p = 16$
$\frac{2p}{2}=\frac{16}{2}$
$p=8$

Is there any difference between the rules used in the two solutions?

Equation Inequality Is the inequality still true?
$2p + 4 = 20$ $2p + 4 < 20$ no
$2p + 4 - 4 = 20 - 4$ $2p+4{\color{red}-4}<20{\color{red}-4}$
$2p = 16$ $2p<16$
$\frac{2p}{2}=\frac{16}{2}$ $\frac{2p}{{\color{red}2}}<\frac{16}{{\color{red}2}}$
$p=8$ $p<8$

No, there is no difference in the rules used for the two solutions.

Example B

In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 6 in for $x$?

Equation Inequality Is the inequality still true?
$3x + 5 = 23$ $3x + 5 \ge 23$ ?
$3x + 5 - 5= 23 - 5$
$3x = 18$
$\frac{3x}{3}=\frac{18}{3}$
$x=6$
Equation Inequality Is the inequality still true?
$3x + 5 = 23$ $3x + 5 \ge 23$ yes
$3x + 5 - 5= 23 - 5$ $3x + 5 {\color{red}-5} \ge 23{\color{red}-5}$
$3x = 18$ $3x\ge 18$
$\frac{3x}{3}=\frac{18}{3}$ $\frac{3x}{{\color{red}3}}\ge \frac{18}{{\color{red}3}}$
$x=6$ $x \ge 6$

No, there is no difference in the rules used for the two solutions.

Example C

In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 3 in for $c$?

Equation Inequality Is the inequality still true?
$5 - 3c = -4$ $5 - 3c \le -4$ ?
$5-5-3c =-4-5$
$-3c = -9$
$\frac{-3c}{-3}=\frac{-9}{-3}$
$c=3$
Equation Inequality Is the inequality still true?
$5 - 3c = -4$ $5 - 3c \le -4$ yes
$5-5-3c =-4-5$ $5{\color{red}-5}-3c \le -4{\color{red}-5}$
$-3c = -9$ $-3c \le -9$
$\frac{-3c}{-3}=\frac{-9}{-3}$ $\frac{-3c}{{\color{red}-3}} \ge \frac{-9}{{\color{red}-3}}$
$c=3$ $c \ge 3$

Yes, there was a difference in the rules used for the two solutions. When dividing by -3, the sign of the inequality was reversed.

Vocabulary

Linear Inequality
Linear inequalities can have one of four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to.

Guided Practice

1. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 10 in for $a$?

Equation Inequality Is the inequality still true?
$4.6a + 8.2 = 2.4a - 13.8$ $4.6a + 8.2 > 2.4a - 13.8$ ?
$4.6a + 8.2+13.8 = 2.4a - 13.8 + 13.8$
$4.6a + 22 = 2.4a$
$4.6a-4.6a + 22 = 2.4a-4.6a$
$22 =-2.2a$
$\frac{22}{-2.2}=\frac{-2.2a}{-2.2}$
$a=10$

2. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 6 in for $w$?

Equation Inequality Is the inequality still true?
$3(w + 4) = 2(3 + 2w)$ $3(w + 4) < 2(3 + 2w)$ ?
$3w + 12 = 6 + 4w$
$3w + 12-12 = 6-12 + 4w$
$3w = -6 + 4w$
$3w-4w = -6 + 4w-4w$
$-w = -6$
$\frac{-w}{-1}=\frac{-6}{-1}$
$w=6$

3. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute -10 in for $h$?

Equation Inequality Is the inequality still true?
$\frac{1}{3}(2-h)=4$ $\frac{1}{3}(2-h) \ge 4$ ?
$\frac{1}{3}(2-h)=4 \left(\frac{3}{3}\right)$
$\frac{1}{3}(2-h)=\frac{12}{3}$
$2-h=12$
$2-2-h=12-2$
$-h=10$
$\frac{-h}{-1}=\frac{10}{-1}$
$h=-10$

1.

Equation Inequality Is the inequality still true?
$4.6a + 8.2 = 2.4a - 13.8$ $4.6a + 8.2 > 2.4a - 13.8$ no
$4.6a + 8.2+13.8 = 2.4a - 13.8 + 13.8$ $4.6a + 8.2{\color{red}+13.8} > 2.4a - 13.8 {\color{red}+13.8}$
$4.6a + 22 = 2.4a$ $4.6a + 22 > 2.4a$
$4.6a-4.6a + 22 = 2.4a-4.6a$ $4.6a{\color{red}-4.6a} + 22 > 2.4a{\color{red}-4.6a}$
$22 =-2.2a$ $22 >-2.2a$
$\frac{22}{-2.2}=\frac{-2.2a}{-2.2}$ $\frac{22}{{\color{red}-2.2}}<\frac{-2.2a}{{\color{red}-2.2}}$
$a=10$ $a<10$

Yes, there was a difference in the rules used for the two solutions. When dividing by -3, the sign of the inequality was reversed.

2.

Equation Inequality Is the inequality still true?
$3(w + 4) = 2(3 + 2w)$ $3(w + 4) < 2(3 + 2w)$ no
$3w + 12 = 6 + 4w$ $3w + 12 = 6 < 4w$
$3w + 12-12 = 6-12 + 4w$ $3w + 12{\color{red}-12} < 6 {\color{red}-12}+4w$
$3w = -6 + 4w$ $3w < -6 + 4w$
$3w-4w = -6 + 4w-4w$ $3w{\color{red}-4w} < -6+4w{\color{red}-4w}$
$-w = -6$ $-w < -6$
$\frac{-w}{-1}=\frac{-6}{-1}$ $\frac{-w}{{\color{red}-1}}<\frac{-6}{{\color{red}-1}}$
$w=6$ $w>6$

Yes, there was a difference in the rules used for the two solutions. When dividing by -3, the sign of the inequality was reversed.

3.

Equation Inequality Is the inequality still true?
$\frac{1}{3}(2-h)=4$ $\frac{1}{3}(2-h) \ge 4$ yes
$\frac{1}{3}(2-h)=4\left(\frac{3}{3}\right)$ $\frac{1}{3}(2-h)=4\left({\color{red}\frac{3}{3}}\right)$
$\frac{1}{3}(2-h)=\frac{12}{3}$ $\frac{1}{3}(2-h)\ge \frac{12}{3}$
$2-h=12$ $2-h \ge 12$
$2-2-h=12-2$ $2{\color{red}-2}-h \ge 12{\color{red}-2}$
$-h=10$ $-h \ge 10$
$\frac{-h}{-1}=\frac{10}{-1}$ $\frac{-h}{{\color{red}-1}} \le \frac{10}{{\color{red}-1}}$
$h=-10$ $h \le-10$

Summary

Linear inequalities are similar to linear equations in that they are relating mathematical expressions. In linear inequalities you use the symbols $>, <, \le$, and $\ge$ rather than the equal sign to show the relationship between the left hand side of the expression and the right had side of the expression.

The rules to solve linear inequalities with one variable are almost the same as those you have learned to solve linear equation with one variable. The difference between these rules is in dividing and multiplying by a negative number. When doing so, you must remember to reverse the sign of the inequality. Other than this, the rules remain the same as you have learned in previous lessons.

It was also found in this lesson that the inequality and the equality have the same solution (or it holds true) if and only if the inequality uses the symbols $\le$ or $\ge$.

Problem Set

1. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute -4.5 in for $x$?
Equation Inequality Is the inequality still true?
$5.2+x+3.6=4.3$ $5.2+x+3.6 \ge 4.3$ ?
$8.8+x=4.3$
$8.8-8.8+x=4.3-8.8$
$x=-4.5$
1. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute 8 in for $n$?
Equation Inequality Is the inequality still true?
$\frac{n}{4}-5=-3$ $\frac{n}{4}-5 <-3$ ?
$\frac{n}{4}-5\left(\frac{4}{4}\right)=-3\left(\frac{4}{4}\right)$
$\frac{n}{4}-\frac{20}{4}=\frac{-12}{4}$
$n-20=-12$
$n-20+20=-12+20$
$n=8$
1. In the following table, a linear equation has been solved. Solve for the inequality using the similar steps. Are the steps the same? Answer the question if the inequality is still true if you substitute -11 in for $z$?
Equation Inequality Is the inequality still true?
$1-z=5(3+2z)+8$ $1-z<5(3+2z)+8$ ?
$1-z=15+10z+8$
$1-z=23+10z$
$1-z+z=23+10z+z$
$1=23+11z$
$1-23=23-23+11z$
$-22=11z$
$\frac{-22}{11}=\frac{11z}{11}$
$z=-11$
1. The sum of two numbers is 764. If one of the numbers is 416, what could the solution be?
2. Two hundred and five less a number is greater than or equal to one hundred and twelve. What could that solution be?

Equation Inequality Is the inequality still true?
$5.2+x+3.6=4.3$ $5.2+x+3.6 \ge 4.3$ yes
$8.8+x=4.3$ $8.8+x \ge 4.3$
$8.8-8.8+x=4.3-8.8$ $8.8{\color{red}-8.8}+x \ge 4.3{\color{red}-8.8}$
$x=-4.5$ $x \ge-4.5$

No, there is no difference in the rules used for the two solutions.

Equation Inequality Is the inequality still true?
$1-z=5(3+2z)+8$ $1-z<5(3+2z)+8$ no
$1-z=15+10z+8$ $1-z<15+10z+8$
$1-z=23+10z$ $1-z<23+10z$
$1-z+z=23+10z+z$ $1-z+z<23+10z+z$
$1=23+11z$ $1<23+11z$
$1-23=23-23+11z$ $1-23<23-23+11z$
$-22=11z$ $-22<11z$
$\frac{-22}{11}=\frac{11z}{11}$ $\frac{-22}{11}<\frac{11z}{11}$
$z=-11$ $z<-11$

No, there is no difference in the rules used for the two solutions.

1. Two hundred and five $\boxed{\text{less}}$ a number is greater than or equal to one hundred and twelve.

$205 - x &\ge 112\\205 {\color{red}- 205} - x &\ge 112 {\color{red}-205}\\-x &\ge -93\\\frac{-x}{-1} & \le \frac{-93}{-1}\\x &\le 93$

Therefore the solution would be any number less than or equal to 93.

## Solving Linear Inequalities in One Variable Algebraically

Introduction

In the previous lessons you had learned about linear relations. Linear relations were of the form $ax + b = 0$, where $a \ne 0$. With linear relations, the mathematical expressions always used the equal sign. Linear inequalities are mathematical statements relating expressions by using one or more inequality symbols $<, >, \le$, or $\ge$. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to.

If you recall at the beginning of this chapter you looked at the method of using a balance to solve algebraic problems. Look at the example below.

To solve for $a + 2 = 5$, you would draw the following balance:

If you were to use the balance method to solve linear inequations, it would look more like this: $a + 2 >5$

Think about it this way. It is like having someone heavier on the $(a + 2)$ side of the seesaw and someone light on the (5) side of the seesaw. The $(a + 2)$ person has a weight greater than (>) the (5) person and therefore the seesaw moves down to the ground.

The rules for solving inequalities are basically the same as you used for solving linear equations. In other words, whatever you do to one side of the equation, you have to do to the other side, even though there is no equal sign, there is now an inequality sign. There are a few additional ones. One big rule is to reverse the sign of the inequality if you are multiplying both sides by a negative number.

It is important for you to remember what the symbols mean. Always remember that the mouth of the sign opens toward the larger number. So $8 >5$, the mouth of the > sign opens toward the 8 so 8 is larger than 5. You know that’s true.

$6b-5 < 300$, the mouth opens toward the 300, so 300 is larger than $6b - 5$. See how it works?

Watch This

Guidance

The Morgan Silver Dollar is a very valuable American dollar minted between 1978 and 1921. When placed on a scale with twenty 1-gram masses, the scale tips toward the Morgan Dollar. Draw a picture to represent this scenario and then write the inequality.

Let $m =$ Morgan dollar

Since the weight of the Morgan dollar is greater than 20 g, the mouth of the inequality sign would open towards the variable, $m$. Therefore the inequality equation would be:

$m>20$

Example A

$15 < 4 + 3x$

To solve variables in inequalities, you need to use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other.

$15 {\color{red}-4} &< 4 {\color{red}-4} + 3x && \approx \text{Subtract 4 from both sides to isolate the variable}\\11 &< 3x && \approx \text{Simplify}\\\frac{11}{{\color{red}3}} &< \frac{3x}{{\color{red}3}} && \approx \text{Divide by 3 to solve for the variable}\\x &> \frac{11}{3} && \approx \text{Simplify}\\$

Let’s just do a quick check to see if this is true. $\frac{11}{3}$ is approximately 3.67. Try substituting 0, 3, and 4 into the equation.

$15 &< 4 + 3x && 15 < 4 + 3x && 15 < 4 + 3x\\15 &< 4 + 3 ({\color{red}0}) && 15 < 4 + 3 ({\color{red}3}) && 15 < 4 + 3 ({\color{red}4})\\15 &< 4 \ \text{(False)} && 15 < 13 \ \text{(False)} && 15 < 16 \ \text{(True)}$

The only value where $x > \frac{11}{3}$ is 4. If you look at the statements above, it was the only inequality that gave a true statement.

Example B

$2y + 3 > 7$

Again, to solve this inequality, use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other.

$2y + 3 {\color{red}-3} &> 7 {\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\2y &> 4 && \approx \text{Simplify}\\\frac{2y}{{\color{red}2}} &> \frac{4}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\y &> 2 && \approx \text{Simplify}$

Let’s just do a quick check to see if this is true. Try substituting 0, 4, and 8 into the equation.

$2y + 3 &> 7 && \ \ 2y + 3 > 7 && \ \ 2y + 3 > 7\\2({\color{red}0}) + 3 &> 7 && 2({\color{red}4}) + 3 > 7 && 2({\color{red}8}) + 3 > 7\\3 &> 7 \ \text{(False)} && \qquad \ 11 > 7 \ \text{(True)} && \qquad \ 19 > 7 \ \text{(True)}$

The values where $y>2$ are 4 and 8. If you look at the statements above, these values when substituted into the inequality gave true statements.

Example C

$-2c-5 < 8$

Again, to solve this inequality, use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other.

$-2c-5{\color{red}+5} &< 8 {\color{red}+5} && \approx \text{Add 5 to both sides to isolate the variable}\\-2c &<13&& \approx \text{Simplify}\\\frac{-2c}{{\color{red}-2}} &< \frac{13}{{\color{red}-2}} && \approx \text{Divide by -2 to solve for the variable}$

Note: When you divide by a negative number, the inequality sign reverses.

$c > \frac{-13}{2} && \approx \text{Simplify}$

Let’s just do a quick check to see if this is true. $\frac{-13}{2}$ is equal to -6.5. Try substituting -8, 0, and 2 into the equation.

$-2c - 5 &< 8 && \quad -2c - 5 < 8 && \quad -2c - 5 < 8\\-2({\color{red}-8}) - 5 &< 8 && -2({\color{red}0}) - 5 < 8 && -2({\color{red}2}) - 5 < 8\\11 &< 8 \ \text{(False)} && \qquad \quad -5 < 8 \ \text{(True)} && \qquad \qquad \ -9 < 8 \ \text{(True)}$

The values where $c>\frac{-13}{2}$ are 0 and 2. If you look at the statements above, these values when substituted into the inequality gave true statements.

Vocabulary

Linear Inequality

Linear inequalities are mathematical statements relating expressions by using one or more inequality symbols $<, >, \le$, or $\ge$. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to.

Guided Practice

1. $4t + 3 > 11$

2. $2z + 7 \le 5z + 28$

3. $9(j - 2) \ge 6(j + 3) - 9$

1. $4t + 3 &> 11\\4t+3{\color{red}-3} &> 11{\color{red}-3}&& \approx \text{Subtract 3 from both sides to isolate the variable}\\4t &> 8&& \approx \text{Simplify}\\\frac{4t}{{\color{red}4}} &< \frac{8}{{\color{red}4}}&& \approx \text{Divide by 4 to solve for the variable}\\t &> 2$

Let’s just do a quick check to see if this is true. Try substituting -2, 0, and 4 into the equation.

$4t + 3 &> 11 && \quad 4t + 3 > 11 && \quad 4t + 3 > 11\\4({\color{red}-2}) + 3 &> 11 && 4({\color{red}0}) + 3 > 11 && 4({\color{red}4}) + 3 > 11\\-5 &> 11 \ \text{(False)} && \qquad \quad 3 > 11 \ \text{(True)} && \qquad \ \ 19 > 11 \ \text{(True)}$

The values where $t > 2$ are 0 and 4. If you look at the statements above, these values when substituted into the inequality gave true statements.

2. $2z + 7 &\le 5z + 28\\2z{\color{red}-2z}+7 &\le 5z{\color{red}-2z}+28 && \approx \text{Subtract} \ 2z \ \text{from both sides to get variables on same side of the inequality sign.}\\7 &\le 3z+28 && \approx \text{Simplify}\\7 {\color{red}-28} &\le 3z+28 {\color{red}-28} && \approx \text{Subtract 28 from both sides to isolate the variable}\\-21 &\le 3z && \approx \text{Simplify}\\\frac{3z}{{\color{red}3}} &\ge \frac{-21}{{\color{red}3}} && \approx \text{Divide by 3 to solve for the variable}\\z &\ge -7$

Let’s just do a quick check to see if this is true. Try substituting -10, 0, and 10 into the equation.

$2z + 7 &\le 5z + 28 && \quad 2z + 7 \le 5z + 28 && \quad \ 2z + 7 \le 5z + 28\\2({\color{red}-10}) + 7 &\le 5({\color{red}-10})+28 && 2({\color{red}0}) + 7 \le 5({\color{red}0}) + 28 && 2({\color{red}10}) + 7 \le 5({\color{red}10}) + 28\\-13 &\le -22 \ \text{(False)} && \qquad \quad 7 \le 28 \ \text{(True)} && \qquad \quad 27 \le 78 \ \text{(True)}$

The values where $z \ge -7$ are 0 and 10. If you look at the statements above, these values when substituted into the inequality gave true statements.

3. $9(j-2) &\ge 6(j + 3)-9\\9j-18 & \ge 6j+18-9 && \approx \text{Remove brackets}\\9j-18 & \ge 6j+9&& \approx \text{Combine like terms on each side of inequality sign}\\9j{\color{red}-6j}-18 & \ge 6j{\color{red}-6j}+9&& \approx \text{Subtract} \ 6j \ \text{from both sides to get variables on same side of the inequality sign.}\\3j-18 & \ge 9&& \approx \text{Simplify}\\3j-18{\color{red}+18} & \ge 9{\color{red}+18}&& \approx \text{Add 18 to both sides to isolate the variable}\\3j &\ge 27&& \approx \text{Simplify}\\\frac{3j}{{\color{red}3}} &\ge \frac{27}{{\color{red}3}}&& \approx \text{Divide by 3 to solve for the variable}\\j &\ge 9$

Let’s just do a quick check to see if this is true. Try substituting -10, 0, and 10 into the equation.

$9(j - 2) &\ge 6(j + 3) - 9 && \quad 9(j - 2) \ge 6(j + 3) - 9 && \quad \ 9(j - 2) \ge 6(j + 3) - 9\\9(({\color{red}-10})-2) &\ge 6(({\color{red}-10}) + 3) - 9 && 9(({\color{red}0}) -2) \ge 6(({\color{red}0}) + 3) - 9 && 9(({\color{red}10}) -2) \ge 6(({\color{red}10}) + 3) -9\\9(-12) &\ge 6(-3) - 9 && \quad \ \ 9(-2) \ge 6(3)-9 && \qquad \quad 9(8) \ge 6(13) - 9\\-108 &\ge -27 \ \text{(False)} && \qquad -18 \ge 9 \ \text{(False)} && \qquad \quad \ \ 72 \ge 69 \ \text{(True)}$

The value where $j \ge 9$ is 10. If you look at the statements above, this values when substituted into the inequality is the only true statement.

Summary

Linear inequalities are very similar to linear equations in that they are mathematical statements relating expressions. The difference between the two is that with linear equations, there is an equal sign relating these expressions. With linear inequations, there is an inequality sign $(>, <, \ge, \le)$.

To solve an inequality, the steps are very similar to the steps for solving linear equations. For example, if you have brackets, remove these by using the distributive property. You must isolate the variable by moving constants to one side and variables to the other side of the inequality sign. You also have to remember that whatever you do to one side of the inequality, you must do to the other. The same was true when you were working with linear equations.

When multiplying and dividing with inequalities, a big point to remember is that when you multiply or divide by a negative number, you must reverse the sign of the inequality. In other words if it is a greater than or equal to sign $(\ge)$ in the inequality, it would change to a less than or equal to sign $(\le)$.

Problem Set

Solve for the variable in the following inequalities.

1. $a+8>4$
2. $4c-1>7$
3. $5-3k<6$
4. $3-4t \le -11$
5. $6 \ge 11-2b$

Solve for the variable in the following inequalities.

1. $\frac{e}{5}-3 >-1$
2. $\frac{1}{5}(r-3) <-1$
3. $\frac{1}{3}(f+2) <4$
4. $\frac{p+3}{4} \ge -2$
5. $\frac{1}{2}(5-w) \le -3$

Solve for the variable in the following inequalities.

1. $3(2x-5)<2(x-1)+3$
2. $2(y+8)+5(y-1)>6$
3. $2(d-3)<-3(d+3)$
4. $3(g+3) \ge 2(g+1)-2$
5. $2(3s-4)+1 \le 3(4s+1)$

Solve for the variable in the following inequalities.

$a+8 &> 4\\a+8{\color{red}-8} &> 4{\color{red}-8} && \approx \text{Subtract eight from both sides to isolate the variable}\\a &> -4 && \approx \text{Simplify}$

Let’s just do a quick check to see if this is true. Try substituting -8, 0, and 4 into the equation.

$a + 8 &> 4 && \quad a + 8 > 4 && \ \ a + 8 > 4\\({\color{red}-8}) + 8 &> 4 && ({\color{red}0})+ 8 > 4 && ({\color{red}4}) + 8 > 4\\0 &> 4 \ \text{(False)} && \qquad \ 8 > 4 \ \text{(True)} && \qquad 12 > 4 \ \text{(True)}$

The values where $a > -4$ are 0 and 4. If you look at the statements above, these values when substituted into the inequality gave true statements.

$5-3k &< 6\\5{\color{red}-5}-3k &< 6{\color{red}-5} && \approx \text{Subtract 5 from each side to isolate the variable}\\-3k &< 1 && \approx \text{Simplify}\\\frac{-3k}{{\color{red}-3}} &> \frac{1}{{\color{red}-3}}&& \approx \text{Divide by -4 to solve for the variable. Remember to reverse the sign of the inequality.}\\k &> \frac{-1}{3}$

Let’s just do a quick check to see if this is true. Try substituting -2, 0, and 4 into the equation.

$5 - 3k &< 6 && \ \ 5-3k < 6 && \ \ 5-3k < 6\\5 - 3({\color{red}-2}) &< 6 && 5-3({\color{red}0}) < 6 && 5-3({\color{red}4}) < 6\\11 &< 6 \ \text{(False)} && \qquad \quad 5 < 6 \ \text{(True)} && \quad \ \ -7 < 6 \ \text{(True)}$

The values where $k > \frac{-1}{3}$ are 0 and 4. If you look at the statements above, these values when substituted into the inequality gave true statements.

$6 &\ge 11-2b\\6{\color{red}-11} &\ge 11{\color{red}-11}-2b && \approx \text{Subtract 11 from both sides to isolate the variable}\\-5 &\ge -2b && \approx \text{Simplify}\\\frac{-5}{{\color{red}-2}} &\le \frac{-2b}{{\color{red}-2}} && \approx \text{Divide by -2 to solve for the variable. Remember to reverse the inequality sign.}\\b &\ge \frac{5}{2}$

Let’s just do a quick check to see if this is true. Try substituting 0, 2, and 4 into the equation.

$6 &\ge 11 - 2b && 6 \ge 11 - 2b && 6 \ge 11 - 2b\\6 &\ge 11 - 2({\color{red}0}) && 6 \ge 11 - 2({\color{red}2}) && 6 \ge 11 - 2({\color{red}4})\\ 6 &\ge 11 \ \text{(False)} && 6 \ge 7 \ \text{(False)} && 6 \ge 3 \ \text{(True)}$

The value where $b \ge \frac{5}{2}$ is 4. If you look at the statements above, this value when substituted into the inequality gave a true statement.

Solve for the variable in the following inequalities.

$\frac{e}{5}-3 &> -1\\\frac{e}{5}-3{\color{red}+3} &> -1{\color{red}+3} && \approx \text{Add 3 to both sides to isolate the variable.}\\\frac{e}{5} &> 2&& \approx \text{Simplify}\\\frac{e}{5}({\color{red}5}) &> 2({\color{red}5})&& \approx \text{Multiply by 5 to solve for the variable}\\e &> 10$

Let’s just do a quick check to see if this is true. Try substituting 0, 10, and 20 into the equation.

$\frac{e}{5}-3 &> -1 && \quad \ \frac{e}{5}-3 > -1 && \quad \ \frac{e}{5}-3 > -1\\\frac{({\color{red}0})}{5}-3 &> -1 && \frac{({\color{red}10})}{5}-3 > -1 && \frac{({\color{red}20})}{5}-3 > -1\\0-3 &> -1 && \quad \ 2-3 >-1 && \quad \ 4-3 >-1 \\-3 &> -1 \ \text{(False)} && \quad \ \ -1 > -1 \ \text{(False)} && \qquad \quad 1>-1 \ \text{(True)}$

The value where $e > 10$ is 20. If you look at the statements above, this value when substituted into the inequality gave a true statement.

$\frac{1}{3}(f+2)&<4\\\frac{1}{3}f+\frac{2}{3}&<4 && \approx \text{Remove brackets}$

This is a LCD problem. The LCD for 3 and 1 is 3, therefore multiply the last number by $\frac{3}{3}$ in order to get a common denominator.

$\frac{1}{3}f+\frac{2}{3} &< 4 \left({\color{red}\frac{3}{3}}\right)\\\frac{1}{3}f+\frac{2}{3} &< \frac{12}{3} && \approx \text{Simplify}$

Since the denominators are all the same, the inequality becomes:

$f+2 &<12\\f+2{\color{red}-2}&<12{\color{red}-12} && \approx \text{Subtract 2 from both sides to isolate and solve for the variable.}\\f &< 10 && \approx \text{Simplify}$

Let’s just do a quick check to see if this is true. Try substituting 0, 10, and 20 into the equation.

$\frac{1}{3}(f+2) &< 4 && \quad \frac{1}{3}(f+2) < 4 && \quad \ \frac{1}{3}(f+2) < 4\\\frac{1}{3}(({\color{red}0})+2) &< 4 && \frac{1}{3}(({\color{red}10})+2) <4 && \frac{1}{3}(({\color{red}20})+2) <4\\\frac{1}{3}(2) &< 4 && \qquad \ \frac{1}{3}(12) < 4 && \qquad \ \ \frac{1}{3}(22) < 4\\\frac{2}{3} &< 4 \ \text{(True)} && \qquad \qquad \ 4>4 \ \text{(False)} && \qquad \qquad \frac{22}{3}<4 \ \text{(False)}$

The value where $f < 10$ is 0. If you look at the statements above, this value when substituted into the inequality gave a true statement.

$\frac{1}{2}(5-w) &\le -3\\\frac{5}{2}-\frac{w}{2} &\le -3 && \approx \text{Remove brackets}\\$

This is a LCD problem. The LCD for 2 and 1 is 2, therefore multiply the last number by $\frac{2}{2}$ in order to get a common denominator.

$\frac{5}{2}-\frac{w}{2} &\le -3 \left({\color{red}\frac{2}{2}}\right)\\\frac{5}{2}-\frac{w}{2} &\le \frac{-6}{2} && \approx \text{Simplify}$

Since the denominators are all the same, the inequality becomes:

$5-w &\le-6\\5{\color{red}-5}-w & \le -6{\color{red}-5} && \approx \text{Subtract 5 from both sides to isolate and solve for the variable.}\\-w &\le -11 && \approx \text{Simplify}\\\frac{-w}{{\color{red}-1}} &\ge \frac{-11}{{\color{red}-1}} && \approx \text{Divide by -1 to solve for the variable. Remember to reverse the inequality sign.}\\w &\ge 11$

Let’s just do a quick check to see if this is true. Try substituting 0, 11, and 20 into the equation.

$\frac{1}{2}(5-w) &\le -3 && \quad \frac{1}{2}(5-w) \le -3 && \quad \frac{1}{2}(5-w) \le -3\\\frac{1}{2}(5-({\color{red}0})) &\le -3 && \frac{1}{2}(5-({\color{red}11})) \le -3 && \frac{1}{2}(5-({\color{red}20})) \le -3\\\frac{1}{2}(5) &\le -3 && \qquad \ \frac{1}{2}(-6) \le -3 && \quad \ \ \frac{1}{2}(-15) \le -3\\\frac{5}{2} &\le -3 \ \text{(False)} && \qquad \quad \ -3 \le -3 \ \text{(True)} && \qquad \quad \frac{-15}{2} \le -3 \ \text{(True)}$

The values where $w \ge 11$ are 11 and 20. If you look at the statements above, these values when substituted into the inequality gave true statements.

Solve for the variable in the following inequalities.

$3(2x-5) &< 2(x-1)+3\\6x-15 &< 2x-2+3 && \approx \text{Remove brackets }\\6x-15 &< 2x+1 && \approx \text{Combine like terms on each side of inequality sign}\\6x{\color{red}-2x}-15 &< 2x{\color{red}-2x}+1 && \approx \text{Subtract} \ 2x \ \text{from both sides to get variables on same side of the inequality sign.}\\4x-15 &< 1 && \approx \text{Simplify}\\4x-15{\color{red}+15} &< 1 {\color{red}+15}&& \approx \text{Add 15 to both sides to isolate the variable}\\4x &< 16&& \approx \text{Simplify}\\\frac{4x}{{\color{red}4}} &< \frac{16}{\color{red}4}&& \approx \text{Divide by 4 to solve for the variable}\\x &< 4$

$2(d-3) &< -3(d+3)\\2d-6 &< -3d+9 && \approx \text{Remove brackets}\\2d{\color{red}+3d}-6 &< -3d{\color{red}+3d}+9 && \approx \text{Add} \ 3d \ \text{to both sides to get variables on same side of the inequality sign.}\\5d-6 &<9 && \approx \text{Simplify}\\5d-6{\color{red}+6} &< 9 {\color{red}+6} && \approx \text{Add 6 to both sides to isolate the variable}\\5d &< 15 && \approx \text{Simplify}\\\frac{5d}{{\color{red}5}} &< \frac{15}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\d &< 3$

$2(3s-4)+1 &\le 3(4s+1)\\6s-24+1 &\le 12s+3 && \approx \text{Remove brackets}\\6s-23 &\le 12s+3 && \approx \text{Combine like terms}\\6s{\color{red}-6s}-23 &\le 12s{\color{red}-6s}+3 && \approx \text{Subtract} \ 6s \ \text{from both sides to get variables on same side of the inequality sign.}\\-23 &\le 6s+3 && \approx \text{Simplify}\\-23{\color{red}-3} &\le 6s+3{\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\-26 &\le 6s && \approx \text{Simplify}\\\frac{6s}{{\color{red}6}} &\ge \frac{-26}{\color{red}6} && \approx \text{Divide by 6 to solve for the variable}\\s &\ge \frac{-13}{3}$

## Graphing the Solution Set of Linear Inequalities in One Variable on a Number Line

Introduction

In the previous two sections you have learned how to find the solution set for linear inequalities with one variable using algebra. After finding the solution set of a linear equation, it is often helpful to visualize the solution on a number line. You have learned some of the steps for using number lines in earlier concepts. In this lesson you will focus on learning how to graph a solution set on a number line for linear inequalities with one variable.

Watch This

Guidance

Jack is 3 years older than his brother. How old are they if the sum of their ages is greater than 17? Write an inequality and solve. Represent the solution set on a number line.

If first let’s write down what you know:

Let $x =$ Jack’s brother’s age

Let $x + 3 =$ Jack’s age

The equation would therefore be:

$x+x+3 &> 17\\2x+3 &> 17 && (\approx \text{Combine like terms})\\2x+3{\color{red}-3} &> 17{\color{red}-3} && (\approx \text{Subtract 3 from both sides to solve for the variable})\\2x &> 14 && (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}} &> \frac{14}{{\color{red}2}} && (\approx \text{Divide by 2 to solve for the variable})\\x &> 7$

Therefore if Jack’s brother is 8 (since $8 > 7$), Jack would be 11.

Representing on a number line

Example A

Represent the solution set to the following inequality on a number line.

$x - 2 &< 5\\x-2{\color{red}+2} &< 5{\color{red}+2} && (\approx \text{Subtract 2 from both sides to isolate the variable})\\x &< 3 && (\approx \text{Simplify})$

Example B

Represent the solution set to the following inequality on a number line.

$3x + 2 &\ge 5\\3x+2{\color{red}-2} &\ge 5{\color{red}-2} && (\approx \text{Subtract 2 from both sides to isolate the variable})\\3x &\ge 3 && (\approx \text{Simplify})\\\frac{3x}{{\color{red}3}} &\ge \frac{3x}{{\color{red}3}} && (\approx \text{Divide by 3 to solve for the variable})\\x &\ge 1$

Example C

Represent the solution set to the following inequality on a number line.

$-3x + 8 &\le 17\\-3x+8{\color{red}-8} &\le 17{\color{red}-8} && (\approx \text{Subtract 8 from both sides to isolate the variable})\\-3x &\le 9 && (\approx \text{Simplify})\\\frac{-3x}{{\color{red}-3}} &\ge \frac{9}{{\color{red}-3}} && (\approx \text{Divide both sides by -3 to solve for the variable, reverse sign of inequality})\\x &\ge -3$

Vocabulary

Number Line
A number line is a line that matches a set of points and a set of numbers one to one.

It is often used in mathematics to show mathematical computations.

Guided Practice

1. Represent the solution set to the inequality $3(a - 1) < 9$ on a number line.

2. Represent the solution set to the inequality $2b + 4 \ge 5b + 19$ on a number line.

3. Represent the solution set to the inequality $0.6c + 2 \ge 5.6$ on a number line.

1. $3(a-1)&<9\\3a-3 &< 9 && (\approx \text{Remove brackets})\\3a-3{\color{red}+3} &< 9 {\color{red}+3} && (\approx \text{Add 3 to both sides to isolate the variable})\\3a &< 12 && (\approx \text{Simplify})\\\frac{3a}{{\color{red}3}} &< \frac{12}{{\color{red}3}} && (\approx \text{Divide both sides by 3 to solve for the variable})\\a &< 4$

2. $2b+4 &\ge 5b+19\\2b{\color{red}-5b}+4 &\ge 5b{\color{red}-5b}+19 && (\approx \text{Subtract} \ 5b \ \text{from both sides to get variables on same side})\\-3b+4 &\ge 19 && (\approx \text{Simplify})\\-3b+4{\color{red}-4} &\ge 19 {\color{red}-4} && (\approx \text{Subtract 4 from both sides to isolate the variable})\\-3b &\ge 15 && (\approx \text{Simplify})\\\frac{-3b}{{\color{red}-3}} &\le \frac{15}{{\color{red}-3}} && (\approx \text{Divide by -3 to solve for the variable, reverse sign of inequality})\\b &\le -5$

3. $0.6 c+2 &\ge 5.6\\0.6c+2{\color{red}-2} &\ge 5.6{\color{red}-2} && (\approx \text{Subtract 2 from both sides to isolate the variable})\\0.6c &\ge 3.6 && (\approx \text{Simplify})\\\frac{0.6c}{{\color{red}0.6}} &\ge \frac{3.6}{{\color{red}0.6}} && (\approx \text{Divide both sides by 0.6 to solve for the variable})\\c &\ge 6$

Summary

Graphing the solution set to an inequality provides a visual representation of the solution. This allows you to “see” all of the numbers that could be possible solutions to the linear inequality. It is important to remember, as you have learned in the past, that when graphing solutions to inequalities, the open circle shows that the number is not included as part of the solution and the closed circle shows that the number is included as part of the solution. The red line, as you have used previously, would show all of the numbers that are possible solutions to the inequality.

Problem Set

1. Represent the solution set to the inequality $-4v > 12$ on a number line.
2. Represent the solution set to the inequality $\frac{-2r}{3} > 4$ on a number line.
3. Represent the solution set to the inequality $4(t-2) \le 24$ on a number line.
4. The prom committee is selling tickets for a fundraiser for the decorations. Each ticket costs $3.50. What is the least number of tickets they need to sell to make$1000? Write an inequality and solve. Graph the solution on a number line.
5. Brenda made 69%, 72%, 81%, and 88% on her last four major tests. How much does she need on her last test to have an average of at least 85% going into the exam? Write an inequality and solve. Graph the solution on a number line.

$-4v &> 12\\\frac{-4v}{\color{red}-4} &< \frac{12}{{\color{red}-4}} && (\approx \text{Divide both sides by -4 to solve for the variable, reverse sign of inequality})\\v &< -3$

$4(t-2) &\le 24\\4t-8 &\le 24 && (\approx \text{Remove brackets})\\4t-8{\color{red}+8} & \le 24{\color{red}+8} && (\approx \text{Add 8 to both sides to isolate the variable})\\4t &\le 32 && (\approx \text{Simplify})\\\frac{4t}{{\color{red}4}} &\le \frac{32}{{\color{red}4}} && (\approx \text{Divide both sides by 4 to solve for the variable})\\t &\le 8$

1. Brenda’s marks: Let $m =$ last mark needed $72\%, 81 \%, 88 \%, 94 \%, m \%$ Brenda wants at least an 85% average The equation would therefore be:

$\frac{1}{5}(72+81+88+94+m) & \ge 85\\\frac{1}{5}(335+m) &\ge 85 && (\approx \text{Combine terms inside brackets to simplify})\\({\color{red}5}) \frac{1}{5}(335+m) &\ge 85({\color{red}5}) && (\approx \text{ Multiply both sides by 5 to remove fraction})\\335+m &\ge 425 && (\approx \text{Simplify})\\335{\color{red}-335}+m &\ge 425{\color{red}-335} && (\approx \text{Subtract 335 from both sides to isolate the variable})\\m &\ge 90$

The line is stopped at 100 because since you are working with percentages (marks), the top mark Brenda could receive is 100%.

## Summary

In this lesson, you began you learning of linear inequalities. Remember that linear inequalities do not have an equal sign but rather us the $>, <, \ge$, and $\le$ signs to relate the two mathematical expressions.

The rules for solving linear inequalities remain the same as for linear equations with one exception. Therefore whatever you do to one side of the inequality sign, you must do to the other side. The exception is if you have to multiply or divide by a negative number. If this happens, you have to reverse the sign of the inequality.

At the end of this lesson, you learned to graph inequalities on a real number line. Graphing the solution set for an inequality works the same as you learned in the previous chapter. You need to use an open circle for > and <. You used a closed circle for $\le$ and $\ge$.

Jan 16, 2013

Jun 04, 2014