(k + 1)x + (k + 2)y + 5 = 0
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Step-by-step explanation:
→ (k + 1)x + (k + 2)y + 5 = 0
→ kx + x + ky + 2y + 5 = 0
→ k(x+y) + x + 2y + 5 = 0
→ lhs ≠ rhs
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