How do you solve number 2 on page 28 with the plugging in method?
Here’s the question (from the Plugging In chapter in the Math Guide):
The way to go here is to plug in values to evaluate choices, which makes this a bit of a hybrid between a plugging in question and a backsolving question. Use the answer choices to guide your plugging in.
For example, choice A tells us to try x < 0. That turns out to be a good idea. Say x = –2 and y equals, oh, I dunno, 3. That fits the conditions: x < y and –3x > y. Can you come up with any values of x greater than 0 where that would work? No? Well, maybe A is correct, then! But just for the moment, let’s try to eliminate the other choices to reassure ourselves.
What about choice B, though? Wouldn’t x = –2 still work if y = –1? Sure would, so we can eliminate that one.
Can we plug in values that would eliminate choice C? Sure can! In fact, x = –2 and y = 3, which we just used above when we were considering choice A, show that choice C doesn’t need to be true.
Does choice D have to be true? Well, it’s true when x = –2 and y = 3, but what if x = –2 and y = 5 instead? Then x < y would be true, and –3x > y would be true, but (–2)^2 is not greater than 5, so choice D doesn’t need to be true.
By trying a few quick numbers informed by the answer choices, we can eliminate B, C, and D, leaving only choice A standing. We’re done!