Return to the Purplemath home page

Try a demo lesson Join Login to


Index of lessons
Print this page (print-friendly version) | Find local tutors


Solving Linear Inequalities:
     Advanced Examples
(page 3 of 3)

Sections: Introduction and formatting, Elementary examples, Advanced examples

  • The velocity of an object fired directly upward is given by V = 80 – 32t, where t is in seconds.

    When will the velocity be between
    32 and 64 feet per second?
  • I will set up the compound inequality, and then solve for t:

    32 < 80 – 32t < 64
    32 – 80 < 80 – 80 – 32t < 64 – 80

    –48 < –32t < –16

    –48 / –32  >  –32t / –32  >  –16 / –32

    1.5 > t > 0.5

    Note that, since I had to divide through by a negative, I had to flip the inequality signs. Note also that you might (as I do) find the above answer to be more easily understood if written the other way around:

      0.5 < t < 1.5

    Looking back at the original question, it did not ask for the value of the variable "t", but asked for the times when the velocity was between certain values. So the actual answer is:

      The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality "0.5 < t < 1.5" did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.


  • Solve 5x + 7 < 3(x + 1).

    First I'll multiply through on the right-hand side, and then solve as usual:

      5x + 7 < 3(x + 1)
      5x + 7 < 3x + 3

      2x + 7 < 3

      2x < –4

      x < –2

Since I divided through by a positive "2" to get the final answer, I didn't have to flip the inequality sign.

  • You want to invest $30,000. Part of this will be invested in a stable 5%-simple-interest rate account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

    First, I have to set up equations for this. The interest formula for simple interest is I = Prt, where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years. Since no time-frame is specified for this problem, I'll assume that t = 1. I'll let "x" be the amount that I'm going to "invest" with my father. Then there will be 30000 – x left to invest in the safe account. The interest on the business investment, assuming that I get paid back, will be:

      (x)(0.07)(1) = 0.07x

    The interest on the safe investment will be:

      (30 000 – x)(0.05)(1) = 1500 – 0.05x

    Then the total interest is:

      0.07x + (1500 – 0.05x) = 0.02x + 1500

    I need to get at least $1900, so:

      0.02x + 1500 > 1900
      0.02x > 400

      x > 20 000

    That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

      I will invest $20,000 at 7%.

  • An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

    This is similar to a mixture word problem, except that this will involve inequality signs, rather than "equals" signs. I'll set it up the same way, though:

  pounds % copper pounds copper
60% x 0.6 0.6x
40% 30 – x 0.4 0.4(30 – x) = 12 – 0.4x
mix 30 between 0.46 and 0.5 between 13.8 and 15

    How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture (30) by the minimum and maximum percentages (46% and 50%, respectively). That is, I multiplied across the bottom row, just as I did in the "60%" row and the "40%" row, to get the right-hand column's value. The total amount of copper in the mixture will be the sum of the copper from the two alloys put into the mixture, so I'll add the expressions for the amount of copper from the alloys, and place the total between the minimum and the maximum allowable amounts of copper:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

      13.8 < 0.6x + (12 – 0.4x) < 15
      13.8 < 0.2x + 12 < 15

      1.8 < 0.2x < 3

      9 < x < 15

    I will need to use between 9 and 15 pounds of the 60% alloy.

  • Solve3(x – 2) + 4 > 2(2x – 3).

    First I'll multiply through and simplify; then I'll solve:

      3(x – 2) + 4 > 2(2x – 3)
      3x – 6 + 4 > 4x – 6
      3x – 2 > 4x – 6

      –2 > x – 6            (*)

      4 > x

      x < 4

Why did I move the "3x" over to the right-hand side (to get to the line marked with a star), instead of moving the "4x" to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided off that coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, "4 > x", is perfectly acceptable. As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

<< Previous  Top  |  1 | 2 | 3  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Solving Linear Inequalities: Advanced Examples." Purplemath. Available from Accessed


   Copyright © 2021  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing


Contact Us