असमानता \(\frac{2x-1}{3}\ge \frac{x+5}{2}\) का सही हल कौन सा है?

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

Explanation opens after your attempt
Correct Answer

A. \(x\ge 17\)

Step 1

Concept

Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.

Step 2

Why this answer is correct

The correct answer is A. \(x\ge 17\). Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.

Step 3

Exam Tip

धनात्मक (6) से गुणा करने पर (2(2x-1)\ge 3(x+5)), इसलिए \(x\ge 17\) है। हर हटाते समय धनात्मक गुणक चिन्ह नहीं बदलता।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \(x\ge 17\). Explanation: धनात्मक (6) से गुणा करने पर (2(2x-1)\ge 3(x+5)), इसलिए \(x\ge 17\) है। हर हटाते समय धनात्मक गुणक चिन्ह नहीं बदलता। / Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.

Which concept should I revise for this Mathematics MCQ?

Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence \(x\ge 17\). A positive multiplier used to clear denominators does not change the sign.

What exam hint can help solve this Mathematics question?

धनात्मक (6) से गुणा करने पर (2(2x-1)\ge 3(x+5)), इसलिए \(x\ge 17\) है। हर हटाते समय धनात्मक गुणक चिन्ह नहीं बदलता।