Question

Write the following expressions in equivalent forms not involving absolute values. (a) \(a+b+|a-b|\) (b) \(a+b-|a-b|\) (c) \(a+b+2 c+|a-b|+|a+b-2 c+| a-b||\) (d) \(a+b+2 c-|a-b|-|a+b-2 c-| a-b||\)

Short answer

Question: Rewrite the expressions without using absolute values: a) \(a+b+|a-b|\) b) \(a+b-|a-b|\) c) \(a+b+2c+|a-b|+|a+b-2c+|a-b||\) d) \(a+b+2c-|a-b|-|a+b-2c-|a-b||\) Answer: a) \(a + b + \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\) b) \(a + b - \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\) c) \(a + b + 2c + \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases} + \begin{cases} a+b-2c+\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr) & \text{if } a+b-2c+|a-b| \ge 0 \\ -(a+b-2c+\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr)) & \text{if } a+b-2c+|a-b| \le 0 \end{cases}\) d) \(a + b + 2c - \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases} - \begin{cases} a+b-2c-\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr) & \text{if } a+b-2c-|a-b| \ge 0 \\ -(a+b-2c-\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr)) & \text{if } a+b-2c-|a-b| \le 0 \end{cases}\)

Step-by-step solution

a) \(a+b+|a-b|\)

We have to consider two cases: 1. If \(a-b \ge 0\), then \(|a-b| = a-b\). Therefore, the expression becomes \(a+b+(a-b)\). 2. If \(a-b \le 0\), then \(|a-b| = -(a-b)\). Hence, the expression becomes \(a+b+(-a+b)\).

Equivalent form for (a)

The equivalent expression for the given case (a) is: \[a + b + \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\]

b) \(a+b-|a-b|\)

We have to look at two separate cases: 1. If \(a-b \ge 0\), then \(|a-b| = a-b\). Therefore, the expression gets simplified to \(a+b-(a-b)\). 2. If \(a-b \le 0\), then \(|a-b| = -(a-b)\). Expression becomes \(a+b+(a-b)\).

Equivalent form for (b)

The equivalent expression for the given case (b) is: \[a + b - \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\]

c) \(a+b+2c+|a-b|+|a+b-2c+|a-b||\)

We will break down this into smaller parts that we can analyze more easily. Let \(p=|a-b|\) and \(q=|a+b-2c+|a-b||\). The expression becomes: \[a+b+2c+p+q\] Now, we have to look at different cases for \(p\) and \(q\) depending on their absolute values: 1. If \(a-b \ge 0\), then \(p = a-b\). 2. If \(a-b \le 0\), then \(p = -(a-b) = b-a\). For the q term: 1. If \(a+b-2c+p \ge 0\), then \(q = a+b-2c+p\). 2. If \(a+b-2c+p \le 0\), then \(q = -(a+b-2c+p)\).

Equivalent form for (c)

The equivalent expression for the given case (c) is: \[a + b + 2c + \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases} + \begin{cases} a+b-2c+\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr) & \text{if } a+b-2c+|a-b| \ge 0 \\ -(a+b-2c+\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr)) & \text{if } a+b-2c+|a-b| \le 0 \end{cases}\]

d) \(a+b+2c-|a-b|-|a+b-2c-|a-b||\)

We will follow the same procedure as in (c). Let \(p=|a-b|\) and \(q=|a+b-2c-|a-b||\). The given expression becomes: \[a+b+2c-p-q\] Now, proceed with cases for \(p\) and \(q\): 1. If \(a-b \ge 0\), then \(p = a-b\). 2. If \(a-b \le 0\), then \(p = -(a-b) = b-a\). For the q term: 1. If \(a+b-2c-p \ge 0\), then \(q = a+b-2c-p\). 2. If \(a+b-2c-p \le 0\), then \(q = -(a+b-2c-p)\).

Equivalent form for (d)

The equivalent expression for the given case (d) is: \[a + b + 2c - \begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases} - \begin{cases} a+b-2c-\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr) & \text{if } a+b-2c-|a-b| \ge 0 \\ -(a+b-2c-\Bigl(\begin{cases} a-b & \text{if } a \ge b \\ b-a & \text{if } a \le b \end{cases}\Bigr)) & \text{if } a+b-2c-|a-b| \le 0 \end{cases}\]

Key concepts

equivalent expressions

When dealing with expressions that involve absolute values, our goal is often to rewrite them as an equivalent expression that doesn't include any absolute values. This makes the expression easier to handle and analyze.
In mathematics, equivalent expressions are different expressions that represent the same value. In the context of absolute values, this usually requires us to consider the underlying conditions that make the absolute value positive or negative.

If you look at the expression \( a + b + |a-b| \), it's clear that the value of the absolute expression \(|a-b|\) depends on whether \(a\) is greater than, equal to, or less than \(b\). Thus, equivalent expressions are found by examining these different cases.

• If \( a - b \ge 0 \), then \(|a-b| = a - b\), leading to an equivalent expression \( a + b + (a-b) \).
• If \( a - b \le 0 \), then \(|a-b| = -(a-b) = b-a\), leading to \( a + b + (-a+b) \).

In this way, equivalent expressions unravel complex pieces into manageable forms depending on different scenarios.

case analysis

Case analysis is a crucial step in simplifying expressions with absolute values. By breaking down the problem into separate cases based on the value of the expression inside the absolute value, we simplify our approach.

The method involves checking all possible conditions under which the expression inside an absolute value could change its sign. For each condition, we calculate and simplify the expression.

• In the expression \( a+b-|a-b| \), the two cases are whether \(a-b\) is positive or negative.
  • Case 1: If \( a - b \ge 0 \), which means \(a\) is greater than or equal to \(b\), the expression simplifies directly to \( a+b-(a-b) \).
  • Case 2: If \( a - b \le 0 \), implying \(a\) is less than \(b\), you replace the absolute value giving \( a + b + (a-b) \).

Case analysis allows us to dissect complex expressions into easier, straightforward statements for each scenario defined by the problem. This analysis is essential when solving for equivalent expressions in algebraic problems.

inequalities

Inequalities play a pivotal role when dealing with absolute values, as they help determine the structure of the equivalent expressions we're working with.
Absolute values inherently involve inequalities since they measure distance from zero, disregarding direction (sign). In an expression like \( |a-b| \), knowing if \( a-b \) is positive or negative tells us whether \( a \) is greater than or less than \( b \). This is primarily established through inequality relationships:

• If \( a \ge b \), then \( a - b \ge 0 \), meaning the absolute value \( |a-b| \) will equal \( a-b \).
• If \( a \le b \), then \( a - b \le 0 \), indicating \( |a-b| = b-a \).

Understanding these inequalities is necessary when simplifying expressions that involve absolute values. It directs us on how to remove the absolute value signs, offering a clearer path to finding equivalent expressions. This logical breakdown helps troubleshoot algebraic expressions comprehensively.