Math Problem Statement

|x - 1| = 2x + 1

Solution

The equation shown in the image is:

x1=2x+1|x - 1| = 2x + 1

This is an absolute value equation, so we must break it down into two cases to solve:

Case 1: Positive case

In this case, we assume x10x - 1 \geq 0, so the absolute value can be removed directly: x1=2x+1x - 1 = 2x + 1 Solve for xx: x1=2x+1    x2x=1+1    x=2    x=2x - 1 = 2x + 1 \implies x - 2x = 1 + 1 \implies -x = 2 \implies x = -2 Check if this solution satisfies the condition x10x - 1 \geq 0: 21=3(which is false)-2 - 1 = -3 \quad (\text{which is false}) Thus, x=2x = -2 is not a valid solution for this case.

Case 2: Negative case

In this case, we assume x1<0x - 1 < 0, so the absolute value must be negated: (x1)=2x+1    x+1=2x+1-(x - 1) = 2x + 1 \implies -x + 1 = 2x + 1 Solve for xx: 11=2x+x    0=3x    x=01 - 1 = 2x + x \implies 0 = 3x \implies x = 0 Check if this solution satisfies the condition x1<0x - 1 < 0: 01=1(which is true)0 - 1 = -1 \quad (\text{which is true}) Thus, x=0x = 0 is a valid solution.

Final Solution:

The only solution to the equation is x=0x = 0.

Would you like more details, or do you have any other questions?

Here are some related questions you can explore:

  1. What are the steps to solve general absolute value equations?
  2. How can we check solutions for absolute value equations?
  3. Can absolute value equations have no solutions? Why or why not?
  4. What happens if the equation inside the absolute value is quadratic?
  5. How can graphing help visualize the solution to this equation?

Tip: When solving absolute value equations, always split into two cases: one for when the inside is non-negative and another for when it’s negative.

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Math Problem Analysis

Mathematical Concepts

Algebra
Absolute Value Equations

Formulas

|a| = a if a >= 0; |a| = -a if a < 0

Theorems

Basic absolute value properties

Suitable Grade Level

Grades 9-10