How do you solve #4x-2(1-x)=2(3x-2)#?

1 Answer
Jun 23, 2016

Answer:

You cannot, it is a contradiction.

Explanation:

Two ways:

1.
#4x - 2(1-x) = 2(3x-2) # factor out 2

#2(2x - (1-x)) = 2(3x-2)# divide by 2

#2x - 1 + x = 3x - 2# simplify

#3x - 1 = 3x -2# + 1 to both sides (by this point we can see it is a contradiction)

# 3x = 3x - 1 # divide by 3

# x ≠ x - (1/3) # impossible / false statment

2.
#4x-2(1-x) = 2(3x-2) # expand brackets

#4x - 2 + 2x = 6x - 4 # combine like terms (simplify)

#6x - 2 = 6x-4# divide by 2 (by this point we can see it is a contradiction)

#3x - 1 = 3x -2# + 1

# 3x = 3x - 1 # divide by 3

# x ≠ x - (1/3) # impossible / false statment

In both ways we see a statement which is false . You can never have #x = x - (1/3)# or any similar statement - it's like saying 1 = 1 - (1/3). It is just not true. Therefore the original statement is a contradiction and cannot be solved.