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2.9: One Variable Inequalities

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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.

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Khan Academy One Step Inequalities

Guidance

One variable inequalities have a different form than one variable equations. Linear equations have the general form of ax + b = c , where a \ne 0 . Linear inequalities can have one of four forms: ax + b > c, ax + b < c, ax + b \ge c, or ax + b \le c . You should notice the difference is that instead of an equals sign, there is an inequality symbol.

When you solve for a linear inequality, you follow the same rules as you would for a linear equation; however, you must remember one big rule: If you divide or multiply by a negative number while solving, the sign of the inequality is reversed.

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.

Concept Problem Revisited

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?

The answer would be yes.

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.

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

Answers:

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

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 -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 -2 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=-2
  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?

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Dec 19, 2012

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Apr 29, 2014
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