असमानता \(\frac{2x+7}{3}\le\frac{x+11}{2}\) का संख्या रेखा पर सही हल क्या है?

What is the correct number line solution of \(\frac{2x+7}{3}\le\frac{x+11}{2}\)?

Explanation opens after your attempt
Correct Answer

A. \(x\le19\), (19) पर बंद बिंदु और बाईं ओर\(x\le19\), closed dot at (19) shaded left

Step 1

Concept

Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.

Step 2

Why this answer is correct

The correct answer is A. \(x\le19\), (19) पर बंद बिंदु और बाईं ओर / \(x\le19\), closed dot at (19) shaded left. Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.

Step 3

Exam Tip

(6) से गुणा करने पर \(4x+14\le3x+33\), इसलिए \(x\le19\)। परीक्षा में भिन्न हटाने के लिए LCM से गुणा करें।

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 \(\frac{2x+7}{3}\le\frac{x+11}{2}\) का संख्या रेखा पर सही हल क्या है? / What is the correct number line solution of \(\frac{2x+7}{3}\le\frac{x+11}{2}\)?

Correct Answer: A. \(x\le19\), (19) पर बंद बिंदु और बाईं ओर / \(x\le19\), closed dot at (19) shaded left. Explanation: (6) से गुणा करने पर \(4x+14\le3x+33\), इसलिए \(x\le19\)। परीक्षा में भिन्न हटाने के लिए LCM से गुणा करें। / Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.

Which concept should I revise for this Mathematics MCQ?

Multiplying by (6) gives \(4x+14\le3x+33\), so \(x\le19\). In exams, multiply by the LCM to clear fractions.

What exam hint can help solve this Mathematics question?

(6) से गुणा करने पर \(4x+14\le3x+33\), इसलिए \(x\le19\)। परीक्षा में भिन्न हटाने के लिए LCM से गुणा करें।