Why do I get different answers when solving equations in different ways?

[member 529879]

when solving 4 = 2(x + 2) + 2, I noticed that only get a correct answer when subtracting the two before dividing the other two.why is this? I'm also confused about solving x - 3 = (x - 3)^2. You can get the right answer by setting it equal to 0. You then get two answers. If you just divide both sides by x - 3 and continue solving, you only get one answer. Why?

 

[Mark44]

when solving 4 = 2(x + 2) + 2, I noticed that only get a correct answer when subtracting the two before dividing the other two.why is this?

If you divide both sides by 2, you get 2 = (x + 2) + 1, so that's a valid step, but a little longer than if you had subtracted 2 from each side as the first step.

I'm also confused about solving x - 3 = (x - 3)^2. You can get the right answer by setting it equal to 0. You then get two answers. If you just divide both sides by x - 3 and continue solving, you only get one answer. Why?

If you bring all terms to one side, you get (x - 3)² - (x - 3) = 0, which I suppose is what you mean by "setting it equal to 0." One way to continue is to factor the left side, which results in (x - 3)[x - 3 - 1] = 0, or (x - 3)(x - 4) = 0. The solutions are x = 3 or x = 4.

The wrong way to do this is to divide by x - 3. If it turns out that x = 3 is a solution, then x - 3 = 0, so you're dividing by zero, which is not allowed. If you do this division, you get x - 3 = 1, or x = 4, so you have lost one of the solutions.