Math Problem Statement

Solve the following absolute value equations and inequalities: 1. |3x| = 6 2. |2x + 5| = 1 Solve the following inequalities: 1. |2x - 3| ≥ 5 2. 1 ≤ |3x + 1| < 5

Solution

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Question 6: Solve the following equations:

(1) 3x=6|3x| = 6

For absolute value equations, we consider two cases:

  1. 3x=63x = 6
  2. 3x=63x = -6

Case 1: 3x=6    x=63=23x = 6 \implies x = \frac{6}{3} = 2

Case 2: 3x=6    x=63=23x = -6 \implies x = \frac{-6}{3} = -2

So, the solution to 3x=6|3x| = 6 is: x=2orx=2x = 2 \quad \text{or} \quad x = -2


(2) 2x+5=1|2x + 5| = 1

Similarly, we consider two cases:

  1. 2x+5=12x + 5 = 1
  2. 2x+5=12x + 5 = -1

Case 1: 2x+5=1    2x=15=4    x=42=22x + 5 = 1 \implies 2x = 1 - 5 = -4 \implies x = \frac{-4}{2} = -2

Case 2: 2x+5=1    2x=15=6    x=62=32x + 5 = -1 \implies 2x = -1 - 5 = -6 \implies x = \frac{-6}{2} = -3

So, the solution to 2x+5=1|2x + 5| = 1 is: x=2orx=3x = -2 \quad \text{or} \quad x = -3


Solve the following inequalities:

(1) 2x35|2x - 3| \geq 5

We again split into two cases:

  1. 2x352x - 3 \geq 5
  2. 2x352x - 3 \leq -5

Case 1: 2x35    2x8    x42x - 3 \geq 5 \implies 2x \geq 8 \implies x \geq 4

Case 2: 2x35    2x2    x12x - 3 \leq -5 \implies 2x \leq -2 \implies x \leq -1

So, the solution to 2x35|2x - 3| \geq 5 is: x4orx1x \geq 4 \quad \text{or} \quad x \leq -1


(2) 13x+1<51 \leq |3x + 1| < 5

This inequality involves a compound inequality, and we treat it as two parts: 13x+11 \leq |3x + 1| and 3x+1<5|3x + 1| < 5.

For the first part, 13x+11 \leq |3x + 1|, we again split into two cases:

  1. 3x+113x + 1 \geq 1
  2. 3x+113x + 1 \leq -1

Case 1: 3x+11    3x0    x03x + 1 \geq 1 \implies 3x \geq 0 \implies x \geq 0

Case 2: 3x+11    3x2    x233x + 1 \leq -1 \implies 3x \leq -2 \implies x \leq \frac{-2}{3}

For the second part, 3x+1<5|3x + 1| < 5, we split it again into two cases:

  1. 3x+1<53x + 1 < 5
  2. 3x+1>53x + 1 > -5

Case 1: 3x+1<5    3x<4    x<433x + 1 < 5 \implies 3x < 4 \implies x < \frac{4}{3}

Case 2: 3x+1>5    3x>6    x>23x + 1 > -5 \implies 3x > -6 \implies x > -2

Now, combining the results:

  • From 13x+11 \leq |3x + 1|, we have x0x \geq 0 or x23x \leq \frac{-2}{3}.
  • From 3x+1<5|3x + 1| < 5, we have 2<x<43-2 < x < \frac{4}{3}.

Thus, the final solution to 13x+1<51 \leq |3x + 1| < 5 is: 2<x23or0x<43-2 < x \leq \frac{-2}{3} \quad \text{or} \quad 0 \leq x < \frac{4}{3}


Would you like more details or further clarification on any of these steps?

Here are some related questions to explore further:

  1. What is the general method to solve absolute value inequalities?
  2. How can compound inequalities be solved more efficiently?
  3. What are the key differences between absolute value equations and inequalities?
  4. How would the solutions change if the inequality symbols were reversed?
  5. Can you express absolute value inequalities graphically?

Tip: When solving absolute value equations, always consider both the positive and negative cases for the expression inside the absolute value!

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Math Problem Analysis

Mathematical Concepts

Algebra
Absolute Value Equations
Inequalities

Formulas

Absolute value equation: |a| = b implies a = b or a = -b
Absolute value inequality: |a| ≥ b implies a ≥ b or a ≤ -b

Theorems

Properties of absolute values
Basic properties of inequalities

Suitable Grade Level

Grades 9-10