X = 1y = 2 and x = − 1y = 4 are the solutions of the equation y = KX + B, and the values of K and B can be obtained

X = 1y = 2 and x = − 1y = 4 are the solutions of the equation y = KX + B, and the values of K and B can be obtained

Substituting x = 1y = 2 and x = − 1y = 4 into the equation y = KX + B, we can get: K + B = 2, ① − K + B = 4, ②, ① - ②: 2K = - 2, the solution is k = - 1; ① + ②: 2B = 6, the solution is b = 3, so k = − 1b = 3