Math Problem Statement

solve the inequality 0 < (2x-1)/(x+1) < 2.

Solution

To solve the inequality 0<2x1x+1<20 < \frac{2x - 1}{x + 1} < 2, we can break it into two inequalities:

  1. 0<2x1x+10 < \frac{2x - 1}{x + 1}
  2. 2x1x+1<2\frac{2x - 1}{x + 1} < 2

Let's solve these step by step.

Step 1: Solve 0<2x1x+10 < \frac{2x - 1}{x + 1}

For this inequality, we need 2x1x+1>0\frac{2x - 1}{x + 1} > 0, which implies that the numerator and the denominator must have the same sign (either both positive or both negative).

Case 1: Both numerator and denominator are positive

  • 2x1>0x>122x - 1 > 0 \Rightarrow x > \frac{1}{2}
  • x+1>0x>1x + 1 > 0 \Rightarrow x > -1

So, for both to be positive, we need x>12x > \frac{1}{2}.

Case 2: Both numerator and denominator are negative

  • 2x1<0x<122x - 1 < 0 \Rightarrow x < \frac{1}{2}
  • x+1<0x<1x + 1 < 0 \Rightarrow x < -1

Thus, for both to be negative, we need x<1x < -1.

So, the solution to the first inequality is: x<1orx>12x < -1 \quad \text{or} \quad x > \frac{1}{2}

Step 2: Solve 2x1x+1<2\frac{2x - 1}{x + 1} < 2

To solve 2x1x+1<2\frac{2x - 1}{x + 1} < 2, we start by isolating 2x1x+1\frac{2x - 1}{x + 1} on one side of the inequality: 2x1x+12<0\frac{2x - 1}{x + 1} - 2 < 0 Simplify by writing the left side with a common denominator: 2x12(x+1)x+1<0\frac{2x - 1 - 2(x + 1)}{x + 1} < 0 Expanding and combining like terms in the numerator: 2x12x2x+1<0\frac{2x - 1 - 2x - 2}{x + 1} < 0 3x+1<0\frac{-3}{x + 1} < 0

For 3x+1<0\frac{-3}{x + 1} < 0, the inequality holds when x+1>0x + 1 > 0, meaning x>1x > -1.

Step 3: Combine Results

Now we combine the results from both inequalities:

  1. From 0<2x1x+10 < \frac{2x - 1}{x + 1}, we found x<1x < -1 or x>12x > \frac{1}{2}.
  2. From 2x1x+1<2\frac{2x - 1}{x + 1} < 2, we found x>1x > -1.

The solution is the intersection of these two results, which is: x>12x > \frac{1}{2}

Final Answer

The solution to the inequality is: x>12x > \frac{1}{2}

Would you like more details on any part of this solution?

Here are some related questions that may help expand your understanding:

  1. How do we solve inequalities involving rational expressions?
  2. What are the conditions for a rational expression to be positive or negative?
  3. How do we combine solutions from multiple inequalities?
  4. Can we solve similar inequalities with quadratic expressions?
  5. What happens if we change the inequality bounds (for example, 0<2x1x+1<30 < \frac{2x - 1}{x + 1} < 3)?

Tip: Always check the domain of the rational expression first, as some values may be undefined due to division by zero.

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Math Problem Analysis

Mathematical Concepts

Inequalities
Rational Expressions
Algebra

Formulas

Rational inequality: a/b < c/d

Theorems

Solving rational inequalities
Sign analysis of rational expressions

Suitable Grade Level

Grades 9-11