# Solve following system of inequalities. x/(2x+1) ≥ 1/4 and 6x/(4x-1) \< 1/2 The answer is x ın (1/8, 1/4)

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**Created:** [Ekim 11, 2026, 8:18ös UTC](https://sorumatik.co/t/solve-following-system-of-inequalities-x-2x-1-1-4-and-6x-4x-1-1-2-the-answer-is-x-in-1-8-1-4/696015 "2026-10-11T20:18:34Z")\
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**Post date:** [Ekim 11, 2026, 8:18ös UTC](https://sorumatik.co/t/solve-following-system-of-inequalities-x-2x-1-1-4-and-6x-4x-1-1-2-the-answer-is-x-in-1-8-1-4/696015/1 "2026-10-11T20:18:34Z")

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Solve following system of inequalities.  
\frac{x}{2x+1} \geq \frac{1}{4} and \frac{6x}{4x-1} \< \frac{1}{2}  
The answer is x ın (\frac{1}{8}, \frac{1}{4})

 ![Solve following system of inequalities.](https://sorumatik.co/uploads/default/original/3X/1/8/18c80d433d6700f6a38c0f2dc6f2998227c9e185.jpeg)

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**Post date:** [Ekim 11, 2026, 8:18ös UTC](https://sorumatik.co/t/solve-following-system-of-inequalities-x-2x-1-1-4-and-6x-4x-1-1-2-the-answer-is-x-in-1-8-1-4/696015/2 "2026-10-11T20:18:56Z")

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## System of Rational Inequalities

💡 **FORMULAS & RULES USED:**

- To solve rational inequalities of the form \frac{P(x)}{Q(x)} \lesseqgtr 0:
  1. Move all terms to one side so the other side is 0.
  2. Find a common denominator to express as a single rational fraction.
  3. Determine the critical points (zeros of the numerator and denominator).
  4. Construct a sign chart to identify the intervals that satisfy the inequality.
  5. The solution to a system of inequalities is the **intersection** of the individual solution sets: S = S\_1 \cap S\_2.

* * *

🧠 **STEP-BY-STEP SOLUTION:**

**Step 1 — Solve the First Inequality**  
Move \frac{1}{4} to the left-hand side and find a common denominator:

\frac{x}{2x+1} \geq \frac{1}{4} \implies \frac{x}{2x+1} - \frac{1}{4} \geq 0

\frac{4x - (2x + 1)}{4(2x + 1)} \geq 0 \implies \frac{2x - 1}{4(2x + 1)} \geq 0

Find the critical points:

- Numerator: 2x - 1 = 0 \implies x = \frac{1}{2}
- Denominator: 2x + 1 = 0 \implies x = -\frac{1}{2} (restriction: x \neq -\frac{1}{2})

Analyze the sign of the quotient \frac{2x - 1}{2x + 1}:

- For x \< -\frac{1}{2}: \frac{(-)}{(-)} \> 0 \quad (+)
- For -\frac{1}{2} \< x \< \frac{1}{2}: \frac{(-)}{(+)} \< 0 \quad (-)
- For x \geq \frac{1}{2}: \frac{(+)}{(+)} \geq 0 \quad (+)

Thus, the solution set for the first inequality is:

S\_1 = \left(-\infty, -\frac{1}{2}\right) \cup \left[\frac{1}{2}, \infty\right)

* * *

**Step 2 — Solve the Second Inequality**  
Move \frac{1}{2} to the left-hand side and combine terms:

\frac{6x}{4x-1} \< \frac{1}{2} \implies \frac{6x}{4x-1} - \frac{1}{2} \< 0

\frac{2(6x) - (4x - 1)}{2(4x - 1)} \< 0 \implies \frac{12x - 4x + 1}{2(4x - 1)} \< 0 \implies \frac{8x + 1}{2(4x - 1)} \< 0

Find the critical points:

- Numerator: 8x + 1 = 0 \implies x = -\frac{1}{8}
- Denominator: 4x - 1 = 0 \implies x = \frac{1}{4} (restriction: x \neq \frac{1}{4})

Analyze the sign of the quotient \frac{8x + 1}{4x - 1}:

- For x \< -\frac{1}{8}: \frac{(-)}{(-)} \> 0 \quad (+)
- For -\frac{1}{8} \< x \< \frac{1}{4}: \frac{(+)}{(-)} \< 0 \quad (-)
- For x \> \frac{1}{4}: \frac{(+)}{(+)} \> 0 \quad (+)

Thus, the solution set for the second inequality is:

S\_2 = \left(-\frac{1}{8}, \frac{1}{4}\right)

* * *

**Step 3 — Find the Intersection of the Solutions**  
The system requires both conditions to hold simultaneously (S = S\_1 \cap S\_2):

S\_1 = \left(-\infty, -\frac{1}{2}\right) \cup \left[\frac{1}{2}, \infty\right)

S\_2 = \left(-\frac{1}{8}, \frac{1}{4}\right)

Notice that:

- The interval S\_2 = \left(-\frac{1}{8}, \frac{1}{4}\right) lies entirely within the interval \left(-\frac{1}{2}, \frac{1}{2}\right).
- Throughout \left(-\frac{1}{2}, \frac{1}{2}\right), the expression \frac{2x-1}{4(2x+1)} is strictly negative, meaning no points in S\_2 satisfy the first inequality.

Therefore:

S\_1 \cap S\_2 = \emptyset

━━━━━━━━━━━━━━━━━━━━━━━━━━━  
✅ **FINAL ANSWER:**

\text{No solution } (\emptyset)

━━━━━━━━━━━━━━━━━━━━━━━━━━━

🎯 **KEY CONCEPTS:**

- **Sign Chart Method:** Rational inequalities must never be cross-multiplied without knowing the sign of the denominator, as multiplying by a negative number reverses the inequality sign.
- **System Intersection:** A value of x is only a solution to the system if it satisfies **all** inequalities in the system simultaneously.

⚠ **COMMON MISTAKES:**

- ❌ **Incorrect Approach (Regarding the note x \in (\frac{1}{8}, \frac{1}{4})):**
  1. A sign error during the simplification of the second inequality (e.g., writing 8x - 1 \< 0 instead of 8x + 1 \< 0) yields the interval \left(\frac{1}{8}, \frac{1}{4}\right).
  2. Forgetting to intersect this result with the first inequality S\_1. For example, testing x = \frac{1}{5} = 0.2 \in \left(\frac{1}{8}, \frac{1}{4}\right) into the first inequality gives:
\frac{0.2}{2(0.2) + 1} = \frac{0.2}{1.4} = \frac{1}{7} \approx 0.143 \not\geq 0.25
which is false.

- ✅ **Correct Approach:** Always test the critical points carefully and take the intersection of all individual solution sets.

* * *

Would you like to solve another rational inequality system to practice setting up sign charts?