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If (x\in\mathbb{Z}) and (-2\leq \frac{x-1}{3}<3), how many integer solutions are there?यदि (x\in\mathbb{Z}) और (-2\leq \frac{x-1}{3}<3) है तो पूर्णांक हलों की संख्या कितनी है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (15)(15)
Multiplying by (3) gives (-6\leq x-1<9), so (-5\leq x<10). The integers from (-5) to (9) are (15) in total.(3) से गुणा करने पर (-6\leq x-1<9), इसलिए (-5\leq x<10) मिलता है। पूर्णांक (-5) से (9) तक हैं, कुल (15)।
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