If , which of the following must be true?
The answer is but I really don't see how we can simplify this to disprove the others.
The question "which of the following must be true?" asks which statement necessarily follows from . Neither x = y nor x = -y follows for all values of x and y. Indeed, for x = 2 and y = -2, the original formula is true, but x = y is false. Similarly, for x = y = 2, the original formula is true, but x = -y is false. The third statement does follow. Indeed, if , then either , which is what we need, or x + y = 0. In the second case, x = -y and therefore again . I'll let you find counterexamples to the last two implications.
I really don't follow either of these explanations, it seems too hocus-pocus to me as if we are working backwards from the choice..using algebra to divide both sides by a single "factor" (either (x + y) or (x^2 - y^2)) I can obtain all of these choices..I still don't see the proof behind this
That is, if either one is true at a given time, why do they not have to be true all the time? That is, x = y or x = -y. Almost like we are proving one and disproving another and then vice versa to disprove the other and saying they dont have to be true??
Well, in this case it is not wrong to work backwards from the choices because the choices are not the solutions to the original equation but rather some arbitrary statements that may follow from it. Regular methods usually describe how to solve an equation, not how to transform it into some other, possibly nonequivalent, equation. Plato wrote the solution to the original equation in post #2.
I stand by my solution because I described a complete proof that implies .
Recall that two equations E1(x) = E2(x) and F1(x) = F2(x) are called equivalent if for every value of x, both equations are either true or both are false. In other words, both equations have the same solutions. We can say that the first equation implies the second one when for every value of x, if the first equation is true, then the second one is true as well. (If the first equation is false for some x, the second equation can be either true or false). Thus, equations are equivalent if both of them imply each other.
It is important to understand the relationship between E(x) = F(x) and E(x) / G(x) = F(x) / G(x) for some expressions E, F and G. The second equation implies the first one. Indeed, for every x, if E(x) / G(x) = F(x) / G(x), then G(x) ≠ 0 and we can multiply both sides by G(x) to get E(x) = F(x). However, the first equation does not imply the second one because for some x we can have E(x) = F(x) and G(x) = 0, so the second equation does not make sense. Therefore, even though you can obtain x + y = 0 from (x + y)(x^2 - y^2) = 0 by dividing both sides by (x^2 - y^2), it does not mean that (x + y)(x^2 - y^2) = 0 implies x + y = 0. Indeed, when x = y, the equation (x + y)(x^2 - y^2) = 0 is true, but the expression by which we divide is zero and the division cannot be performed, which makes the result of the division x + y = 0 false (in general). Thus, by dividing an equation by an expression containing x you may loose solutions.
I should have made it clear that I was answering this assuming it was in the form of a test-prep question. I happened to recognize this question from either the American ACT or SAT, I forget which. And yes we do train students to work backwards from the distractors towards the answer in test-prep courses. I can remember the very moment I learned that lesson. I was part of a seminar, almost forty years ago, led by a well-known educational tester. He put forth this idea. I objected at once and loudly. He looked at me and said, “I don’t care if they understand the mathematics or not, I care about how well they do on this test”. We became good friends. In fact, he got me into editing test questions.