This activity is intended to supplement Algebra I, Chapter 6, Lesson 6.
Probmem 1 - Introduction to Disjunction and Conjunction
Consider the equation |x|=5. To solve, you would graph both sides of the equation as functions (y=|x| and y=5) and mark the solution as the area where the graphs intersect.
The same method can be applied to inequalities.
Press APPS and select the Inequalz app. Press any key to begin.
Example 1: |x|<5
- Using Y1= graph the left side as y=|x|. The absolute value function is located by pressing MATH → and selecting abs(.
- Using Y2= graph the right side as y<5. On the equals sign, press ALPHA [F2] for the < sign. Press ZOOM and select ZoomStandard.
- Find the intersection points by pressing 2nd [TRACE] and selecting intersect. Now just move the cursor to the intersection point and press ENTER three times. The solution is where the shading overlaps the graph of the absolute value function.
In this case, the solution is −5<x<5.
When an absolute value is less than a number, it is a conjunction because the solution is just one part of the graph.
Example 2: |x−4|≥8
- Using Y1= graph the left side as y=|x−4|.
- Using Y2= graph the right side as y≥8. On the equals sign, press ALPHA[F5] for the ≥ sign. Press WINDOW to choose appropriate window settings.
- Find the intersection points.
In this case, the solution is x≤−4 or x≥12.
When an absolute value is greater than a number, it is a disjunction because the solution is two separate parts of the graph.
Problem 2 - Application of Disjunction and Conjunction
For the problems below, write the inequalities as either a conjunction or disjunction, then solve for x. Check your solution by graphing using the method described in Examples 1 and 2. Please use your graphing calculator to check your results.
Problem 3 - Real World Application
One application of absolute value inequalities is engineering tolerance. Tolerance is the idea that an ideal measurement and an actual measurement can only differ within a certain range.
A bolt with a 10 mm diameter has a tolerance range of 9.965 mm to 10 mm, while the hole that it fits into has a tolerance range of 10.05 mm to 10.075 mm.
How can you express the tolerances of both the bolt and the hole in terms of an absolute value inequality?