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This absolute value inequalities lesson uses the weight gain formula as an example

Subjects:

Math, P.E. & Health  

Grade:

10  

Title – Absolute Value Inequalities
By – Glen Cotant
Primary Subject – Math
Secondary Subject – PE & Health
Grade Level – 10

Standard:

    CA STD 1.0
Total Recall (Warm-up): (5 minutes approx.)
      Exercise from yesterday’s lesson on inequalities.
      Solve for x and graph the solution for this inequality:

200 < 2000 – 0.5x < 500

        (A: 3600

>

        x

>

      3000 )
Direct Instruction: (10 minutes approx.)
      The inequality in today’s “Total Recall” is a formula for

caloric intake with a desire for a weight gain

      .

200 < 2000 – 0.5x < 500

2000

        is the amount of calories taken in (

what you eat

        ). This varies.

0.5x

        represents a half hour of some exercise ‘x’ burning off calories.

        • Brisk jogging ( running ) burns off 800 calories per hour.
        • Walking burns off 500 calories per hour.
        • What exercise would burn 3000 to 3600 calories per hour?
          (None; maximum healthy exercise is around 800 to 1000 calories per hour.)

200 to 500

        is the range of the final daily caloric outcome, and since it is positive, it represents a weight gain.

Q:

      What would have to change for this to become a weight loss situation?

Q:

      What different types of workouts could you add to the list?

Q:

    What if you performed two different workouts during the day, one for ½ hour and one for ¾ hour? How would you express that in an inequality?
Direct Instruction: (10 minutes approx.)
      Another way to describe a calorie gain between 200 to 500 calories would be:

350 ± 150 calories

      The ideal is the middle of the range, and the tolerance is the variation. Here the ideal is 350 and the tolerance is 150. Using absolute value, there is an inequality to describe our initial compound inequality:

| (2000 – 0.5x) – 350 | < 150

      Let’s break this down and see if it is the same as our initial compound inequality. The difference of the ACTUAL and IDEAL can be a positive or a negative value, because of the absolute value symbol.
      Therefore, we have two cases:

        1.

{(2000 – 0.5x) – 350} is a negative value

          (

the absolute value will change it to a positive value, but taking away the absolute value – its going to be a negative value

          ):

-{(2000 – 0.5x) – 350} < 150

          and if we multiply both sides by -1, the inequality direction changes and we have:

{(2000 – 0.5x) – 350} > -150

        2.

{(2000 – 0.5x) – 350} is a positive value

          (

the absolute value doesn’t change it so we end up with

          ):

{(2000 – 0.5x) – 350} < 150

      Putting these both together in a compound inequality:

-150 < {(2000 – 0.5x) – 350} < 150

      and if we add 350 to both sides of the inequality, we get:

200 < (2000 – 0.5x) < 500

    which is what we started with!
Practice: (5 minutes approx.)
      The format:

|ACTUAL – IDEAL| < TOLERANCE

      is used extensively in quality control for production manufacturing. In a bolt making factory, the size of the bolt ideally is 1/4″ and each bolt is measured for accuracy to be up to plus or minus 0.005″ or it is rejected. Write an absolute value inequality for the testing of these bolts:

|x – 0.25| < .005

      where x is the measure of each bolt.
Direct Instruction: (5 minutes approx.)
      Here is a summary of the various absolute value inequalities: (

Copy in notebooks

    ).
 
  Abs. Value Ineq.
Compound Form
Graph of Solution
  |ax + b| < c
  -c < ax + b < c
 
Example: |3x + 5| <10
-10 < 3x + 5 < 10
-5 < x < 5/3       |ax + b| < c
  -c < ax + b < c
 
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