Let​ a, b,​ c, and x be real numbers.
a. How is solving StartAbsoluteValue ax EndAbsoluteValue plus b equals c different from solving StartAbsoluteValue ax plus b EndAbsoluteValue equals c​?
b. How is solving StartAbsoluteValue ax EndAbsoluteValue plus b less than or equals c different from solving StartAbsoluteValue ax plus b EndAbsoluteValue greater than or equals c​?
Question content area bottom
Part 1
a. How is solving StartAbsoluteValue ax EndAbsoluteValue plus b equals c different from solving StartAbsoluteValue ax plus b EndAbsoluteValue equals c​?
A.
When solving StartAbsoluteValue ax EndAbsoluteValue plus b equals c​, b has to be subtracted from both sides first to isolate the absolute value expression.
Your answer is correct.B.
Solving StartAbsoluteValue ax plus b EndAbsoluteValue equals c involves solving two equations that do not have absolute values. Solving StartAbsoluteValue ax EndAbsoluteValue plus b equals c involves solving only one such equation.
C.
Solving StartAbsoluteValue ax EndAbsoluteValue plus b equals c involves solving two equations that do not have absolute values. Solving StartAbsoluteValue ax plus b EndAbsoluteValue equals c involves solving only one such equation.
D.
When solving StartAbsoluteValue ax plus b EndAbsoluteValue equals c​, b has to be subtracted from both sides first to isolate the absolute value expression.
Part 2
b. How is solving StartAbsoluteValue ax EndAbsoluteValue plus b less than or equals c different from solving StartAbsoluteValue ax plus b EndAbsoluteValue greater than or equals c​?

### Part 1
**a. How is solving \(|ax| + b = c\) different from solving \(|ax + b| = c\)?**

**Correct Answer:** A.  
When solving \(|ax| + b = c\), \(b\) has to be subtracted from both sides first to isolate the absolute value expression.

- **Explanation:** In the equation \(|ax| + b = c\), you first subtract \(b\) from both sides to isolate the absolute value, resulting in \(|ax| = c - b\). From there, you solve for \(x\). On the other hand, in \(|ax + b| = c\), the absolute value expression is already isolated, so you immediately consider the two cases where \(ax + b = c\) and \(ax + b = -c\) to solve for \(x\).

### Part 2
**b. How is solving \(|ax| + b \leq c\) different from solving \(|ax + b| \geq c\)?**

**Explanation:**
- For \(|ax| + b \leq c\), you first need to subtract \(b\) from both sides to isolate the absolute value term, giving \(|ax| \leq c - b\). After this, you consider the case where \(- (c - b) \leq ax \leq c - b\).
  
- For \(|ax + b| \geq c\), the absolute value expression is already isolated, and you handle it by considering the two cases:
  1. \(ax + b \geq c\)
  2. \(ax + b \leq -c\)

These lead to two separate inequalities for \(x\).

### Follow-Up Questions:
1. What are the general steps for solving an inequality involving an absolute value?
2. How would the solution change if \(c\) were negative in \(|ax + b| = c\)?
3. What does the solution look like when \(c = 0\) in \(|ax + b| = c\)?
4. How do you interpret the results when the absolute value inequality has no solution?
5. How would you solve the equation if the absolute value term is greater than a constant?

**Tip:** Always ensure the absolute value expression is isolated before breaking it into the two corresponding equations or inequalities.

Let a, b, c, and x be real numbers. How is solving |ax| + b = c different from solving |ax + b| = c? How is solving |ax| + b ≤ c different from solving |ax + b| ≥ c?

Explore the differences between solving absolute value equations like |ax| + b = c and |ax + b| = c. Learn how to handle inequalities such as |ax| + b ≤ c and |ax + b| ≥ c. This guide is perfect for students in Grades 9-11 looking to strengthen their algebra skills.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation