असमानता \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\) का हल क्या है?

What is the solution of \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\)?

Explanation opens after your attempt
Correct Answer

C. \(x\le -\frac{1}{5}\)

Step 1

Concept

Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).

Step 2

Why this answer is correct

The correct answer is C. \(x\le -\frac{1}{5}\). Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).

Step 3

Exam Tip

हर हटाने पर \(9-3x\ge 2x+10\) मिलता है। इसलिए \(-5x\ge 1\) और \(x\le -\frac{1}{5}\) है।

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Mathematics Answer, Explanation and Revision Hints

असमानता \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\) का हल क्या है? / What is the solution of \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\)?

Correct Answer: C. \(x\le -\frac{1}{5}\). Explanation: हर हटाने पर \(9-3x\ge 2x+10\) मिलता है। इसलिए \(-5x\ge 1\) और \(x\le -\frac{1}{5}\) है। / Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).

Which concept should I revise for this Mathematics MCQ?

Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).

What exam hint can help solve this Mathematics question?

हर हटाने पर \(9-3x\ge 2x+10\) मिलता है। इसलिए \(-5x\ge 1\) और \(x\le -\frac{1}{5}\) है।