Math Problem Statement

|2x| + |3 - 2x| ≤ 3

Solution

The inequality in the image is:

2x+32x3|2x| + |3 - 2x| \leq 3

This inequality involves absolute values, so we need to handle it by considering different cases based on the points where the expressions inside the absolute values change sign.

Step-by-Step Solution:

Case 1: x32x \geq \frac{3}{2}

In this case:

  • 2x=2x|2x| = 2x since 2x2x is positive for x32x \geq \frac{3}{2}.
  • 32x=2x3|3 - 2x| = 2x - 3 because 32x3 - 2x becomes negative when x32x \geq \frac{3}{2}.

So, the inequality becomes: 2x+(2x3)32x + (2x - 3) \leq 3 Simplifying: 4x334x - 3 \leq 3 4x64x \leq 6 x32x \leq \frac{3}{2} But this case is for x32x \geq \frac{3}{2}, so the only solution for this case is x=32x = \frac{3}{2}.

Case 2: 0x<320 \leq x < \frac{3}{2}

In this case:

  • 2x=2x|2x| = 2x since 2x2x is positive for x0x \geq 0.
  • 32x=32x|3 - 2x| = 3 - 2x because 32x3 - 2x is positive when x32x \leq \frac{3}{2}.

The inequality becomes: 2x+(32x)32x + (3 - 2x) \leq 3 Simplifying: 2x+32x32x + 3 - 2x \leq 3 333 \leq 3 This is always true, so x[0,32]x \in [0, \frac{3}{2}] is a solution for this case.

Case 3: x<0x < 0

In this case:

  • 2x=2x|2x| = -2x because 2x2x is negative for x<0x < 0.
  • 32x=32x|3 - 2x| = 3 - 2x because 32x3 - 2x is still positive for x<0x < 0.

The inequality becomes: 2x+(32x)3-2x + (3 - 2x) \leq 3 Simplifying: 2x+32x3-2x + 3 - 2x \leq 3 4x+33-4x + 3 \leq 3 4x0-4x \leq 0 x0x \geq 0 But this case is for x<0x < 0, so there is no solution in this case.

Final Solution:

The solution to the inequality is x[0,32]x \in [0, \frac{3}{2}].


Would you like further details on any step, or do you have any questions?

Here are some related questions you might explore:

  1. How do you solve inequalities involving absolute values?
  2. What are the conditions under which absolute value expressions change sign?
  3. How do you handle systems of inequalities?
  4. What is the graphical representation of this inequality?
  5. Can this problem be solved using a different method, like algebraic manipulation?

Tip: When solving absolute value inequalities, always consider the points where each absolute value expression changes sign, and break the problem into cases based on those points.

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Math Problem Analysis

Mathematical Concepts

Algebra
Inequalities
Absolute Value

Formulas

|a| = a if a ≥ 0, |a| = -a if a < 0

Theorems

Properties of Absolute Values
Piecewise Function Representation

Suitable Grade Level

Grades 9-11