Question 704504
You let/force a = b. 


So because a = b, this means that 


a^2 - ab


b^2 - b*b ... plug in a = b (ie replace all copies of 'a' with 'b')


b^2 - b^2


0


So if a = b, then a^2 - ab is equal to 0. 


When you jump from 2(a^2-ab)= 1(a^2-ab) to 2 = 1 in the last two lines, you are dividing both sides by a^2-ab. But if a = b, then a^2-ab = 0, which means you're dividing both sides by zero...which is undefined.


So the last step is NOT a valid algebraic move because you can't divide by zero.


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Thanks,


Jim


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