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When will (\alpha x+4y=7) and (9x+12y=21) have a unique solution?युग्म (\alpha x+4y=7) और (9x+12y=21) का अद्वितीय हल कब होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\alpha\neq3)(\alpha\neq3)
For a unique solution, (\frac{\alpha}{9}\neq\frac{4}{12}) must hold. Thus (\alpha\neq3) is correct.अद्वितीय हल के लिए (\frac{\alpha}{9}\neq\frac{4}{12}) होना चाहिए। इसलिए (\alpha\neq3) सही शर्त है।
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