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INEQUALITIES IN CONTEXT

Unit Overview
In unit 12, you learned to graph a linear inequality. In this unit, we will expand upon that knowledge and put that in context (word problems).  You will also learn to identify constraints on the domain and range based on the information in the problem.


Linear Inequalities in One Variable


To review inequalities, click here.  (Unit 12, Linear Inequalities)



Example #3: Click on the link to watch the video "Writing One Variable Inequalities" or click on the video.

 

After watching this video, we can solve these inequalities to find what the number could be.


The first example gave 2/3(x) – 5 ≥ 11


Add 5 to both sides gives 2/3(x) ≥ 16


Multiply both sides by 3/2 to get x ≥ 16 (3/2)


x  ≥ 24  This means that x must be at least 24 (24 or greater).



Example #4: Click on the link to watch the video "Constructing and solving a one-step inequality" or click on the video.

 

Now, let's take this example further. If x < 333 1/3 tiles, what is the most number of tiles he can purchase? He cannot buy a partial tile, so 333 is the most he can buy. When graphed, the graph should not exceed past the point where x = 333. What is the least number of tiles he could purchase? Well, zero, of course. But then, he wouldn't get a patio!


What is our domain and range for this problem?  The domain is represented by the number of tiles. How many tiles can the contractor purchase?  He can buy anywhere from 0 to 333 tiles. We write this as


0 ≤ x ≤  333.


The range is represented by how much the contractor is spending. He will spend anywhere from $0 to how much 333 tiles cost.  Since 333 × 3 = 999 tiles, 999 is the upper limit of the range. We write this as


0 ≤ y ≤ 999.

 


Stop!  Go to Questions #1-10 about this section, then return to continue on to the next section.

 


Linear Inequalities in Two Variables





Example #2: Click on the link to watch the video "Two Variable Linear Inequality Word Problems" or click on the video.


In this example, what is the domain and range? The domain is found on the x-axis.  This is the number of dolls that can be purchased.  Because the most number of dolls that can be purchased is the x-intercept, this can be found by substituting 0 in for f (no outfits purchased means the most dolls that can be purchased).


100d + 25(0) ≤ 1000


100d ≤ 1000


d ≤ 10


The domain for this problem is anywhere from 0 to 10 dolls. This is written as 0 ≤ d ≤ 10.


The range represents the number of outfits purchased. Again, the maximum is given on the y-axis, so we can find this by substituting in 0 for d (no dolls purchased means the most number of outfits that can be purchased).


10(0) + 25f ≤ 1000


               25f ≤ 1000


                  f ≤ 40


Of course, this was already found when the intercepts were found to make the graph. The range is therefore, 0 ≤ f ≤ 40.


Example #3: Click on the link to watch the video " Linear Inequality in Two Variables Application Problem (Phone Cost: Day and Night)" or click on the video.

 

Now that you have watched this example, find the domain and range of this inequality.  Since the horizontal access was d, this will be our domain.  What are the lowest and highest values for this inequality? Notice the graph starts at the d-intercept (normally you would think of this as your x-intercept but the variable d has been used in this problem) of 50, therefore, x must be 50 or more.  Remember that the problem asked about being charged more than $10, so low values would not make sense.  Therefore, our domain is d ≥ 50.   The night minutes are graphed on the y-axis but are labeled as n for this problem. This is our range. Notice that the graph starts at 200 night minutes so our range is n ≥ 200.

Stop! Go to Questions #11-30 to complete this unit.



Below are additional educational resources and activities for this unit.
 
Variable and Verbal Expressions
 
Graphing Linear Inequalities