3x + k - 5 = kx - k + 13x - kx = - 2k + 6x.(3 - k) = - 2k + 6x = (- 2k + 6)/(3 - k)x = (2k - 6)/(k - 3) → where: k ≠ 3x = 0(2k - 6)/(k - 3) = 02k - 6 = 02k = 6k = 3 → no possible because the conditionx = 1(2k - 6)/(k - 3) = 12k - 6 = k - 32k - k = 6 - 3k = 3 → no possible because the conditionx = 2(2k - 6)/(k - 3) = 22k - 6 = 2.(k - 3)2k - 6 = 2k - 60 = 0 → the solution is: x = 23x + k - 5 = kx - k + 1 → when: x = 26 + k - 5 = 2k - k + 11 + k = k + 1k = k ← okx² = x ← ok, it's possiblex² - x = 0 ← okx.(x - 1) = 0x = 0/(x - 1) ← you can make this operation only if: (x - 1) ≠ 0 → x ≠ 1…but, at the beginning, you said that: x = 1, so the reasoning is no possible.

X^2 - X = 0

X( X - 1 ) = 0

para una X = 0

la otra X = 1

cualquier valor de k (menos el 3), hace que x sea 2

si x es 1, entonces 0/(x - 1) es 0/0?