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Using interval notation to express all real numbers less than or equal to a Or greater than or equal to b

Write the interval expressing all real numbers less than or equal to −1 or greater than or equal to 1.

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at and ends at −1 , which is written as ( , −1 ] .

The second interval must show all real numbers greater than or equal to 1 , which is written as [ 1 , ) . However, we want to combine these two sets. We accomplish this by inserting the union symbol, , between the two intervals.

( , −1 ] [ 1 , )
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Express all real numbers less than −2 or greater than or equal to 3 in interval notation.

( , −2 ) [ 3 , )

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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.

Properties of inequalities

A d d i t i o n   P r o p e r t y If  a < b ,  then  a + c < b + c . M u l t i p l i c a t i o n   P r o p e r t y If  a < b  and  c > 0 ,  then  a c < b c . If  a < b  and  c < 0 ,  then  a c > b c .

These properties also apply to a b , a > b , and a b .

Demonstrating the addition property

Illustrate the addition property for inequalities by solving each of the following:

  • (a) x 15 < 4
  • (b) 6 x 1
  • (c) x + 7 > 9

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. x 15 < 4 x 15 + 15 < 4 + 15   Add 15 to both sides . x < 19

  2. 6 x 1 6 + 1 x 1 + 1 Add 1 to both sides . 7 x

  3. x + 7 > 9 x + 7 7 > 9 7 Subtract 7 from both sides . x > 2
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Solve: 3 x −2 < 1.

x < 1

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Demonstrating the multiplication property

Illustrate the multiplication property for inequalities by solving each of the following:

  1. 3 x < 6
  2. −2 x 1 5
  3. 5 x > 10

  1. 3 x < 6 1 3 ( 3 x ) < ( 6 ) 1 3 x < 2

  2. 2 x 1 5 2 x 6 ( 1 2 ) ( 2 x ) ( 6 ) ( 1 2 ) Multiply by  1 2 . x 3 Reverse the inequality .

  3. 5 x > 10 x > 5 ( 1 ) ( x ) > ( 5 ) ( 1 ) Multiply by  1. x < 5 Reverse the inequality .
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Solve: 4 x + 7 2 x 3.

x −5

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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.

Solving an inequality algebraically

Solve the inequality: 13 7 x 10 x 4.

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

13 7 x 10 x 4 13 17 x −4 Move variable terms to one side of the inequality . −17 x −17 Isolate the variable term . x 1 Dividing both sides by  −17  reverses the inequality .

The solution set is given by the interval ( , 1 ] , or all real numbers less than and including 1.

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Solve the inequality and write the answer using interval notation: x + 4 < 1 2 x + 1.

( 2 , )

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Solving an inequality with fractions

Solve the following inequality and write the answer in interval notation: 3 4 x 5 8 + 2 3 x .

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

3 4 x 5 8 + 2 3 x 3 4 x 2 3 x 5 8 Put variable terms on one side . 9 12 x 8 12 x 5 8 Write fractions with common denominator . 17 12 x 5 8 x 5 8 ( 12 17 ) Multiplying by a negative number reverses the inequality . x 15 34

The solution set is the interval ( , 15 34 ] .

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Questions & Answers

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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
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Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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Practice Key Terms 4

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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