असमानता \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\) का हल क्या है?

What is the solution of the inequality \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\)?

Explanation opens after your attempt
Correct Answer

C. \(x\ge \frac{26}{7}\)

Step 1

Concept

Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.

Step 2

Why this answer is correct

The correct answer is C. \(x\ge \frac{26}{7}\). Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.

Step 3

Exam Tip

हर हटाने पर \(7x-14\ge 12\), अतः \(x\ge \frac{26}{7}\)। भिन्नों में पहले लघुत्तम समापवर्त्य से गुणा करें।

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असमानता \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\) का हल क्या है? / What is the solution of the inequality \(\frac{5x-1}{3}-\frac{x+4}{2}\ge 2\)?

Correct Answer: C. \(x\ge \frac{26}{7}\). Explanation: हर हटाने पर \(7x-14\ge 12\), अतः \(x\ge \frac{26}{7}\)। भिन्नों में पहले लघुत्तम समापवर्त्य से गुणा करें। / Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.

Which concept should I revise for this Mathematics MCQ?

Clearing denominators gives \(7x-14\ge 12\), hence \(x\ge \frac{26}{7}\). For fractions, multiply by the LCM first.

What exam hint can help solve this Mathematics question?

हर हटाने पर \(7x-14\ge 12\), अतः \(x\ge \frac{26}{7}\)। भिन्नों में पहले लघुत्तम समापवर्त्य से गुणा करें।