समीकरणों (0.2x+0.5y=3.1) और (0.4x-0.1y=1.3) से (x+y) का मान क्या है?

What is the value of (x+y) from (0.2x+0.5y=3.1) and (0.4x-0.1y=1.3)?

Explanation opens after your attempt
Correct Answer

A. \(x+y=\frac{97}{11}\)

Step 1

Concept

Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).

Step 2

Why this answer is correct

The correct answer is A. \(x+y=\frac{97}{11}\). Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).

Step 3

Exam Tip

दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)।

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

समीकरणों (0.2x+0.5y=3.1) और (0.4x-0.1y=1.3) से (x+y) का मान क्या है? / What is the value of (x+y) from (0.2x+0.5y=3.1) and (0.4x-0.1y=1.3)?

Correct Answer: A. \(x+y=\frac{97}{11}\). Explanation: दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)। / Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).

Which concept should I revise for this Mathematics MCQ?

Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).

What exam hint can help solve this Mathematics question?

दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)।