Can someone explain why this equation is not true? 1=2?

Question

x=2
x(x-1)=2(x-1)
x^2-x=2x-2
x^2-2x=x-2
x(x-2)=x-2
x=1

clearly x cannot equal 1 and 2 (thus 1=2).

Im not entirly certain why this works out this way, but I assume that multiplying the equation by (x-1) changed something but I cannot figure out what.

Answer 1

This proof, and all it's variants assume (somewhat indirectly) that division by zero is valid as you cancel out the (x-2) before stating x=1

Multiplying by (x-1) changes nothing because the argument is flawed from the beginning in the sense that you're manipulation involves equating zeros.

Hope this helps, ever heard of the proof that 1+2+3+....= -1/12?

That one is true in an odd twisted way...

Answer 2

x^2-x=2x-2
x^2-3x+2=0 this is a 2nd degree equation ( can have more than one solution )
(x-2)(x-1)=0
x=2 and x=1
In fact when you write this x=2 is the same like you write this x-2=0 and when you write x(x-1)=2(x-1) is like you're writing (x-2)(x-1)=0(x-1) that is (x-2)(x-1)=0

Answer 3

I think you have gone wrong here
x=2
x(x-1)=2(2-1)=4-2=2

Answer 4

2x-2 is NOT equal to x^2-x

2x-2 = 4 but

x^2-x =
= -2x OR x =
= -4 OR 2

that means x is 2 and not -4 as 2 isn't equal to -4

x(x-2) = x-2 =>

=> x^2-3x+2=0 => x=2 (Pythagorean Theorem)

there is only ONE correct answer which is x=2

Answer 5

first step
x(x-1)=2(x-1) <== here you introduce a new solution : x = 1 if you dont want that you should say for X <> 1

last step you divide by 0 if x = 2

x(x-2)=x-2
HERE you should say that x = 2 is a solution
NOW continue for x not equal 2

Answer 6

x^2-x=2x-2
x^2-x+x=2x-2+x
x^2=3x-2
x^2-3x+2=0
(x-1)(x-2)=0
x does equal 1 and 2 because it is simplified to a quadractic equation

Answer 7

This equation has more than one root

If you divide both sides by (x-1) in the beginning, you would get 2

This is a polynomial, therefore it has two answers

Hope I helped

Answer 8

b/c in the last step, you can't divide anything by x-2, b/c x-2=2-2=0, and dividing by 0 is undefined.

Answer 9

Actually it can. In step 4 you have a quadratic equation which will yield two answers.