Math Problem Statement

Solve the following inequalities: (a) 2x + 1 > 5 (b) 3x - 8 <= 13 (c) 3 - 2/x > x (d) 1/(2x) <= 7/(16x) (e) x/(2x - 3) <= 2 (f) x/3 - x/(2x) >= 1

Solution

The image contains six inequalities to solve. Let's solve them one by one.

(a) 2x+1>52x + 1 > 5

  1. Subtract 1 from both sides: 2x>42x > 4
  2. Divide by 2: x>2x > 2

(b) 3x8133x - 8 \leq 13

  1. Add 8 to both sides: 3x213x \leq 21
  2. Divide by 3: x7x \leq 7

(c) 32x>x3 - \frac{2}{x} > x

  1. Rearrange: 3>x+2x3 > x + \frac{2}{x}
  2. Multiply both sides by xx (assuming x>0x > 0 because of division): 3x>x2+23x > x^2 + 2 This simplifies to: x23x+2<0x^2 - 3x + 2 < 0
  3. Factor the quadratic: (x1)(x2)<0(x - 1)(x - 2) < 0
  4. From the sign analysis, the solution is: 1<x<21 < x < 2

(d) 12x716x\frac{1}{2x} \leq \frac{7}{16x}

  1. Multiply both sides by 16x16x (assuming x>0x > 0): 878 \leq 7 This is a contradiction, so no solution exists for this inequality.

(e) x2x32\frac{x}{2x - 3} \leq 2

  1. Multiply both sides by 2x32x - 3 (considering the sign of 2x32x - 3): x2(2x3)x \leq 2(2x - 3) This simplifies to: x4x6x \leq 4x - 6
  2. Subtract xx from both sides: 03x60 \leq 3x - 6
  3. Add 6: 63x6 \leq 3x
  4. Divide by 3: 2x2 \leq x
  5. Now, check the domain. Since 2x3>02x - 3 > 0, x>32x > \frac{3}{2}, so the solution is: x2x \geq 2

(f) x3x2x1\frac{x}{3} - \frac{x}{2x} \geq 1

  1. Simplify the second term on the left: x2x=12\frac{x}{2x} = \frac{1}{2} So, the inequality becomes: x3121\frac{x}{3} - \frac{1}{2} \geq 1
  2. Add 12\frac{1}{2} to both sides: x332\frac{x}{3} \geq \frac{3}{2}
  3. Multiply by 3: x92x \geq \frac{9}{2} So the solution is: x4.5x \geq 4.5

Summary of Solutions:

(a) x>2x > 2
(b) x7x \leq 7
(c) 1<x<21 < x < 2
(d) No solution
(e) x2x \geq 2
(f) x4.5x \geq 4.5

Would you like further explanations or details on any step?

Here are some related questions you might consider:

  1. How do you solve inequalities with quadratic expressions?
  2. How do sign changes affect inequality solutions when multiplying or dividing?
  3. How do we handle inequalities with rational expressions?
  4. Can you explain the interval notation for inequality solutions?
  5. What happens when an inequality has no solution?

Tip: When solving inequalities with variables in the denominator, always remember to consider restrictions on the domain to avoid division by zero.

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Math Problem Analysis

Mathematical Concepts

Inequalities
Algebraic manipulation
Quadratic expressions
Rational expressions

Formulas

Factorization of quadratic expressions
Sign analysis for inequalities
Basic algebraic rules for manipulating inequalities

Theorems

Sign changes when multiplying or dividing inequalities
Quadratic inequality solutions

Suitable Grade Level

Grades 9-12