✔ 最佳答案
x = 2 - 1/ (1+yz)
x(1+yz) = 2(1 +yz) - 1 .... [把所以數x(1+yz), - 1/(1+yz) x (1+yz) 會cancel out = -1]
x + xyz = 2 + 2yz - 1 .... [拆括號]
xyz - 2yz = 2 - 1 - x .... [將同y聯繫既放喺左邊, 其他轉去右邊]
y(xz-2z) = 1 - x .... [抽出y, common factor]
y = (1-x)/(xz-2z) .... [將括號搬去右邊做分母]
y = (1-x)/(z(x-2)) .... [因為分母2個數都有z, 所以要抽出z]
xz= 1 - 1/1-y
xz(1-y) = 1(1-y) - 1 .... [把所以數x(1-y), - 1/(1-y) x (1-y) 會cancel out = -1]
xz - xyz = 1 - y - 1
-xyz + y = -xz
-xyz -(y)(-1) = -xz
-y (xz - 1)= -xz
y(xz - 1) = xz .... [因為2邊都係負數, 所以可以刪除左個-號]
y = xz/(xz-1)