To consider the similarities and differences between them, we will solve the given equation and inequalities. Then we can compare them.
| x -1 | = 2
To solve an equation of the form | A | = b, we need to solve two cases, the positive case and the negative case.
lA=b A=- b ⇒ lcx-1=2 & (I) x-1=-2 & (II)
Let's start with the first case.
To solve an inequality in the form | A | ≤ b, we need to solve an "and" compound inequality. This is because the distance from the midpoint must be less than or equal to 2 units away.
- b ≤ A ≤ b ⇒ - 2 ≤ x - 1 ≤ 2
The solution set corresponds to the overlapping region of the two solutions sets for the individual inequalities.
- 2 ≤ x - 1 and x - 1 ≤ 2
Let's work out the first inequality.
The solution for the compound inequality are all those values greater than or equal to - 1 andless than or equal to 3.
| x -1 | ≥ 2
To solve an inequality in the form | A | ≥ b, we need to solve an "or" compound inequality. This is because we need the distance from the midpoint needs to be greater than or equal to 2 units away.
A≤ - b or A≥ b
⇓
x - 1 ≤ - 2 or x - 1 ≥ 2
The solution set contains both of the solution sets of the individual inequalities. Let's work out the first one.
The solutions to the compound inequality are all those numbers less than or equal to -1 or all numbers greater than or equal to 3.
Comparison and Conclusions
Let's summarize the differences and similarities encountered.
Similarities
Each of the situations required us to consider two cases, one using the value 2 the other using its opposite value, - 2. This is a consequence of having the absolute value function in the equation or inequality.
Differences
Each situation was different in terms of what we were solving.
The equation required us to solve two equations.
The inequalities required to solve compound inequalities.
We can also notice how different they are by looking at their solutions sets.
