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प्रश्न
Solve and graph the solution set of:
x + 5 ≥ 4(x – 1) and 3 – 2x < –7, x ∈ R
योग
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उत्तर
Given: x + 5 ≥ 4(x – 1) and 3 – 2x < –7
⇒ x + 5 ≥ 4x – 4
4x – x ≤ 5 + 4
3x ≤ 9
Dividing both sides by 3 we get,
x ≤ 3 ...(i)
3 – 2x < –7
2x > 3 + 7
2x > 10
x > 5 ...(ii)
From (i) and (ii) we get,
x ≤ 3 and x > 5
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