असमानता \(3-\frac{x}{2}\le\frac{2x+9}{6}<5\) का हल क्या है?
What is the solution of \(3-\frac{x}{2}\le\frac{2x+9}{6}<5\)?
Explanation opens after your attempt
A. \(x\ge\frac{3}{5}\) और \(x<\frac{21}{2}\)\(x\ge\frac{3}{5}\) and \(x<\frac{21}{2}\)
Concept
The left inequality gives \(18-3x\le2x+9\), hence \(x\ge\frac{9}{5}\); the right gives \(x<\frac{21}{2}\). The correct intersection is \(x\ge\frac{9}{5}\) and \(x<\frac{21}{2}\).
Why this answer is correct
The correct answer is A. \(x\ge\frac{3}{5}\) और \(x<\frac{21}{2}\) / \(x\ge\frac{3}{5}\) and \(x<\frac{21}{2}\). The left inequality gives \(18-3x\le2x+9\), hence \(x\ge\frac{9}{5}\); the right gives \(x<\frac{21}{2}\). The correct intersection is \(x\ge\frac{9}{5}\) and \(x<\frac{21}{2}\).
Exam Tip
बाईं असमानता से \(18-3x\le2x+9\), अतः \(x\ge\frac{9}{5}\) मिलता है; दाईं से \(x<\frac{21}{2}\)। सही प्रतिच्छेद \(x\ge\frac{9}{5}\) और \(x<\frac{21}{2}\) है।
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