SG #1 Algebra Chapter 7 Medium Questions #4

[johnhurford]

xy < 0 & a/x > b/y.

QA is ay & QB is bx.

I've read the answer explanation and don't understand how if you multiply a/x > b/y by xy how you get to ay < bx? Also can you please show me how plugging in numbers would work as a solution methodology?

Thanks,

John

[tommywallach]

Hey John,

Looks like you need to go back to your fractions! Remember when you multiply a fraction by something, you're only multiplying the NUMERATOR by that something.

a/x * xy = axy/x = ay

b/y * xy = bxy/y = bx

So if you multiply both sides of the given inequality by xy, you get ay on one side and bx on the other. And because you know xy is negative, you know you have to flip the sign.

As for plugging in, it's a really bad call here. In general, you don't want to plug in if you're going to have to set more than 2 variables. Even 3 is plausible (though it would be really time consuming). But if you have to plug in FOUR variables, then test multiple solutions to make sure the answer isn't D? Nah. Nobody can do that in a timely fashion.

This is one of those algebra only type of questions.

Hope that helps!

-t