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Solving Linear Inequalities:
     Elementary Examples (page 2 of 3)

Sections: Introduction and formatting, Elementary examples, Advanced examples

  • Solve 2x < 9.

    If they had given me "2x = 9", I would have divided the 2 from each side. I can do the same thing here:

      x <= 9/2

    Then the solution is: x < 9/2

    ...or, if you prefer decimals (and if your instructor will accept decimal equivalents instead of fractions):

      x < 4.5

  • Solve x/4 > 1/2.

    If they had given me " x/4 = 1/2 ", I would have multiplied both sides by 4. I can do the same thing here:

      x > 2

    Then the solution is: x > 2

  • Solve –2x < 5.

    Remember how I said that solving linear inequalities is "almost" exactly like solving linear equations? Well, this is the one place where it's different. To explain what I'm about to do, consider the following:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

      3 > 2

    What happens to the above inequality when I multiply through by –1? The temptation is to say that the answer will be "–3 > –2". But –3 is not greater than –2; it is in actuality smaller. That is, the correct inequality is actually the following:

      –3 < –2

    As you can see, multiplying by a negative ("–1", in this case) flipped the inequality sign from "greater than" to "less than". This is the new wrinkle for solving inequalities:

      When solving inequalities, if you multiply or divide through by a negative, you must also flip the inequality sign.

    To solve "–2x < 5", I need to divide through by a negative ("–2"), so I will need to flip the inequality:

      '-2x less than 5' transforms itself into 'x greater than -5/2' by dividing through by -2 and flipping the inequality sign

    Then the solution is: x > –5/2

  • Solve (2x – 3)/4  < 2.

    First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

      (2x – 3)/4   < 2
      (4) × (2x – 3)/4  < (4)(2)
      2x – 3 < 8
      2x < 11
      x < 11/2  = 5.5

  • Solve 10 < 3x + 4 < 19.

    This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when I go to subtract the 4, I will have to subtract it from all three "sides".

      10 < 3x + 4 < 19
      6 < 3x < 15
      2 < x < 5

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Cite this article as:

Stapel, Elizabeth. "Solving Linear Inequalities: Elementary Examples." Purplemath. Available from
    http://www.purplemath.com/modules/ineqlin2.htm. Accessed
 

 

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