Modulus Equations

There is quite a lot of confusion on how to solve modulus equations and inequalities. I believe this arises because there are different ways of solving the same problem. Students may have seen these different methods and wonder why some are presented while others are not. Furthermore, sometimes an apparent `solution` does not satisfy the original equation, Overall, the student is either free to choose a method or sometimes the question directs the student in a certain direction for example, a student may be asked to sketch the graph.

Let`s look at some examples:

Method 2: Sketch a graph of each modulus equation and determine where the lines intersect. Check as in method 1.

Method 3: 2x - 1 = x - 2 results in x = -1 and -2x + 1 = x - 2 results in x = 1. Check as in method 1.

x - 1 = 4 - 2x so x = 5/3. Check: 2(5/3) + 1= 13/3 and 5 - |5/3 - 1| = 5 - 2/3 = 13/3. So x = 5/3 is a solution.

-x + 1 = 4 -2x so x = 3. A check shows that x = 3 is not in fact a solution.

Conclusion:

Choose a method to sole the inequality by choice or as guided in the question.

Solve the equation

Check the solution works by replacement in the original equations.