समीकरण (x-2y=-4) और (3x+y=11) का हल कौन-सा है?

Which is the solution of (x-2y=-4) and (3x+y=11)?

Explanation opens after your attempt
Correct Answer

B. बिंदु (\left\(3,2\right\))Point (\left\(3,2\right\))

Step 1

Concept

At (\left\(3,2\right\)), (3-2\left\(2\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{18}{7},\frac{23}{7}\right\)).

Step 2

Why this answer is correct

The correct answer is B. बिंदु (\left\(3,2\right\)) / Point (\left\(3,2\right\)). At (\left\(3,2\right\)), (3-2\left\(2\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{18}{7},\frac{23}{7}\right\)).

Step 3

Exam Tip

(\left\(3,2\right\)) पर (3-2\left\(2\right\)=-1) है इसलिए यह नहीं है। सही हल (\left\(\frac{18}{7},\frac{23}{7}\right\)) है।

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Mathematics Answer, Explanation and Revision Hints

समीकरण (x-2y=-4) और (3x+y=11) का हल कौन-सा है? / Which is the solution of (x-2y=-4) and (3x+y=11)?

Correct Answer: B. बिंदु (\left\(3,2\right\)) / Point (\left\(3,2\right\)). Explanation: (\left\(3,2\right\)) पर (3-2\left\(2\right\)=-1) है इसलिए यह नहीं है। सही हल (\left\(\frac{18}{7},\frac{23}{7}\right\)) है। / At (\left\(3,2\right\)), (3-2\left\(2\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{18}{7},\frac{23}{7}\right\)).

Which concept should I revise for this Mathematics MCQ?

At (\left\(3,2\right\)), (3-2\left\(2\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{18}{7},\frac{23}{7}\right\)).

What exam hint can help solve this Mathematics question?

(\left\(3,2\right\)) पर (3-2\left\(2\right\)=-1) है इसलिए यह नहीं है। सही हल (\left\(\frac{18}{7},\frac{23}{7}\right\)) है।