Résolvons dans ℝ : 3k -14 = 3·(k -5) + 1 3k -14 -3·(k -5) -1 = 0 3k -15 -3·(k -5) = 0 3k -15 -(3k -15) = 0 3k -15 -3k + 15 = 0 ℹ-15 + 15 = 0‖3k -3k = 0 0 + 0 = 0 0 = 0 ▶Solution trouvée : k ∈ ℝ
