असमानता \(\frac{2x+3}{5}-\frac{x-4}{2}\le\frac{7-x}{10}\) का हल क्या है?

What is the solution of \(\frac{2x+3}{5}-\frac{x-4}{2}\le\frac{7-x}{10}\)?

Explanation opens after your attempt
Correct Answer

A. \(x\ge\frac{7}{2}\)

Step 1

Concept

Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.

Step 2

Why this answer is correct

The correct answer is A. \(x\ge\frac{7}{2}\). Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.

Step 3

Exam Tip

धनात्मक (10) से गुणा करने पर (2(2x+3)-5(x-4)\le7-x) मिलता है। सरलीकरण से \(26-x\le7-x\), जो असत्य है, अतः कोई हल नहीं।

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Mathematics Answer, Explanation and Revision Hints

असमानता \(\frac{2x+3}{5}-\frac{x-4}{2}\le\frac{7-x}{10}\) का हल क्या है? / What is the solution of \(\frac{2x+3}{5}-\frac{x-4}{2}\le\frac{7-x}{10}\)?

Correct Answer: A. \(x\ge\frac{7}{2}\). Explanation: धनात्मक (10) से गुणा करने पर (2(2x+3)-5(x-4)\le7-x) मिलता है। सरलीकरण से \(26-x\le7-x\), जो असत्य है, अतः कोई हल नहीं। / Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.

Which concept should I revise for this Mathematics MCQ?

Multiplying by positive (10) gives (2(2x+3)-5(x-4)\le7-x). Simplification gives \(26-x\le7-x\), which is false, so no solution exists.

What exam hint can help solve this Mathematics question?

धनात्मक (10) से गुणा करने पर (2(2x+3)-5(x-4)\le7-x) मिलता है। सरलीकरण से \(26-x\le7-x\), जो असत्य है, अतः कोई हल नहीं।