Math Problem Statement

Solve the following equations/inequalities: 1. (1/x) = 2 / (3x - 1), 2. (7 - 2x) / (x + 3) ≤ 5, 3. (1/x) + (1/2x) = 2/3, 4. (x - 3)/(x + 4) ≤ 2, 5. (1/(x^2 + 4)) + (1/(x^2 - 4)) = 18/(x^4 - 16)

Solution

Let's go through each of the problems step-by-step. I'll start with the first one:

1. 1x=23x1\frac{1}{x} = \frac{2}{3x - 1}

Step 1: Cross-multiply to eliminate the fractions: 1(3x1)=2x1 \cdot (3x - 1) = 2 \cdot x This simplifies to: 3x1=2x3x - 1 = 2x

Step 2: Solve for xx: 3x2x=13x - 2x = 1 x=1x = 1

Solution for Problem 1: x=1x = 1


2. 72xx+35\frac{7 - 2x}{x + 3} \leq 5

Step 1: Multiply both sides by x+3x + 3 (assuming x+30x + 3 \neq 0):

72x5(x+3)7 - 2x \leq 5(x + 3)

Step 2: Distribute the 5 on the right-hand side: 72x5x+157 - 2x \leq 5x + 15

Step 3: Bring all terms involving xx to one side and constants to the other side: 7155x+2x7 - 15 \leq 5x + 2x 87x-8 \leq 7x

Step 4: Solve for xx: x87x \geq -\frac{8}{7}

Solution for Problem 2: x87x \geq -\frac{8}{7}


3. 1x+12x=23\frac{1}{x} + \frac{1}{2x} = \frac{2}{3}

Step 1: Combine the fractions on the left-hand side: 1x+12x=2+12x=32x\frac{1}{x} + \frac{1}{2x} = \frac{2 + 1}{2x} = \frac{3}{2x}

Step 2: Set this equal to the right-hand side: 32x=23\frac{3}{2x} = \frac{2}{3}

Step 3: Cross-multiply: 33=22x3 \cdot 3 = 2 \cdot 2x 9=4x9 = 4x

Step 4: Solve for xx: x=94x = \frac{9}{4}

Solution for Problem 3: x=94x = \frac{9}{4}


4. x3x+42\frac{x - 3}{x + 4} \leq 2

Step 1: Move all terms to one side: x3x+420\frac{x - 3}{x + 4} - 2 \leq 0

Step 2: Combine into a single fraction: x32(x+4)x+40\frac{x - 3 - 2(x + 4)}{x + 4} \leq 0 x32x8x+40\frac{x - 3 - 2x - 8}{x + 4} \leq 0 x11x+40\frac{-x - 11}{x + 4} \leq 0

Step 3: Solve the inequality x11x+40\frac{-x - 11}{x + 4} \leq 0. The critical points are at x=4x = -4 and x=11x = -11.

Analyzing the sign of the expression around these points:

  • For x<11x < -11, the expression is positive.
  • For 11<x<4-11 < x < -4, the expression is negative.
  • For x>4x > -4, the expression is positive.

So, the solution is 11x4-11 \leq x \leq -4 (but x4x \neq -4 since that makes the denominator zero).

Solution for Problem 4: 11x<4-11 \leq x < -4


5. 1x2+4+1x24=18x416\frac{1}{x^2 + 4} + \frac{1}{x^2 - 4} = \frac{18}{x^4 - 16}

Step 1: Recognize that x416=(x2+4)(x24)x^4 - 16 = (x^2 + 4)(x^2 - 4), so the right-hand side simplifies to: 18(x2+4)(x24)\frac{18}{(x^2 + 4)(x^2 - 4)}

Step 2: Combine the left-hand side into a single fraction: (x24)+(x2+4)(x2+4)(x24)=2x2(x2+4)(x24)\frac{(x^2 - 4) + (x^2 + 4)}{(x^2 + 4)(x^2 - 4)} = \frac{2x^2}{(x^2 + 4)(x^2 - 4)}

Step 3: Set this equal to the right-hand side: 2x2(x2+4)(x24)=18(x2+4)(x24)\frac{2x^2}{(x^2 + 4)(x^2 - 4)} = \frac{18}{(x^2 + 4)(x^2 - 4)}

Step 4: Cancel the denominators and solve: 2x2=182x^2 = 18 x2=9x^2 = 9 x=±3x = \pm 3

Solution for Problem 5: x=3x = 3 or x=3x = -3


Would you like more details on any step or have any other questions? Here are five questions to deepen your understanding:

  1. How do you determine the critical points in inequalities with rational expressions?
  2. Why is it important to check for excluded values in rational equations?
  3. How does cross-multiplying help in solving fractional equations?
  4. How can you simplify complex rational expressions before solving them?
  5. What is the significance of analyzing signs in rational inequalities?

Tip: When solving inequalities, always analyze the critical points carefully, and check which intervals satisfy the inequality.

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Math Problem Analysis

Mathematical Concepts

Algebra
Inequalities
Rational Equations
Cross-multiplication
Quadratic Expressions

Formulas

Cross-multiplication
Rational expressions simplification
Solving inequalities using critical points

Theorems

Cross-multiplication method
Solving inequalities by sign analysis

Suitable Grade Level

Grades 9-12