Question

given an equation of the form "num1+num2=nim3" where one of num1, num2, num3 is replaced by an x. find the value of x

Answer 1

Consider the given equation num 1+num 2 = num 3

where one of num1, num2, num3 is replaced by an x.

Case 1: Let us suppose num 1 = x

Therefore, the given equation is:

x + num 2 = num 3

So, x = num 3 - num 2

Therefore, the value of x is "num 3 - num 2".

Case 2: Let us suppose num 2 = x

Therefore, the given equation is:

num 1 + x = num 3

So, x = num 3 - num 1

Therefore, the value of x is "num 3 - num 1".

Case 3: Let us suppose num 3 = x

Therefore, the given equation is:

num 1 + num 2 = x

So, x = num 1 + num 2

Therefore, the value of x is "num 1 + num 2".

Answer 2

If It is given that the equation will definitely be in this form then :

1) '+' sign exist btw num1 and num2.

2) '=' sign exist btw num2 and num3.

Based on this the code will be :

str1 = input("enter equation: ")

num1='p'

num2='q'

num3='r'

for i in range(len(str1)):

   if str1[i]=='+':

       posi_plus = i

       if str1[i-1] == 'x':

           num1 = 'x'

       elif str1[i+1] == 'x':

           num2 = 'x'

       else:

           num3 = 'x'

   

   if str1[i] == '=':

       posi_eq = i

       break

a = str1[:posi_plus ]

b = str1[posi_plus+1 : posi_eq ]

c = str1[posi_eq+1:]

if num1 == 'x':

   print(int(c)-int(b))

elif num2 == 'x':

   print(int(c)-int(a))

else:

   print(int(a)+int(b))

Note : I have not added the corner test cases like x+2x=10,x=0 etc ; This code is just for the basic and T(n) test cases.