如果要迭代以查找满足条件的值a，则需要for-循环或while-循环，并在此循环中更改值a

a = 10.0
b = 5.0
c = 15.0
d = 0.5
e = 1.0

# target values

contaminant_1_target = 10.0
contaminant_2_target = 2.0

step = -0.1  # precision

while True:
    # pseudo outputs
    contaminant_1 = a * b + c
    contaminant_2 = a * d + e
    #print(contaminant_1, contaminant_2)

    if contaminant_1 < contaminant_1_target and contaminant_2 < contaminant_2_target:
        print(a)
        break
    else:
        a += step

编辑：

但如果你使用了错误的step，那么它可能会计算在错误的方向，它将永远运行。它需要检查，如果当前值大于以前的值，并作出更正。
它看起来类似于machine learning和linear regresion，它们使用Least squares来查找匹配函数
它将(contaminant_1 - contaminant_1_target)^2与先前的值进行比较，以查看它是否必须改变方向。

编辑：

比较当前和以前的(contaminant_1 - contaminant_1_target)^2并更改step = -step的版本-它可以从低于或高于结果的a开始。
（维基百科：least squares）

# - constants -  `UPPER_CASE_NAMES`

B = 5.0
C = 15.0
D = 0.5
E = 1.0

# - variables - `lower_case_names`

a = 1.0

# a = 10.0

# --------------------------------

# target values

contaminant_1_target = 10.0
contaminant_2_target = 2.0

step = -0.1  # precision

# at start it should be rather `infinity`

previous_LS_1 = 0  # least squares
previous_LS_2 = 0  # least squares

while True:
    # pseudo outputs
    contaminant_1 = a * B + C
    contaminant_2 = a * D + E
    #print(contaminant_1, contaminant_2)

    # ---

    current_LS_1 = (contaminant_1 - contaminant_1_target)**2
    current_LS_2 = (contaminant_2 - contaminant_2_target)**2

    if current_LS_1 > previous_LS_1 and current_LS_2 > previous_LS_2:
        print('change direction:', a)
        step = -step

    previous_LS_1 = current_LS_1
    previous_LS_2 = current_LS_2

    # ---

    if contaminant_1 < contaminant_1_target and contaminant_2 < contaminant_2_target:
        print('a:', a)
        break
    else:
        a += step