Is x|y|>y^2?

[cgcivello]

Is x|y| > y^2 ?

1) x > y
2) y > 0




...


_____________
Answer = C

The explanation states that both are needed because that answer can be yes or no depending on whether x and y are positive or negative. The explanation also explicitly warns you of dividing by |y| (and eventually getting to 'Is x > y?') because you do not know the value of y. This is where I am confused... since the absolute value of a number is always positive, isn't it okay to divide by both sides by it and keeping the inequality sign facing the same way? While y might be positive or negative, |y| will always be positive, so why can you do this?

[jlucero]

Is x|y| > y^2 ?

1) x > y
2) y > 0




...


_____________
Answer = C

The explanation states that both are needed because that answer can be yes or no depending on whether x and y are positive or negative. The explanation also explicitly warns you of dividing by |y| (and eventually getting to 'Is x > y?') because you do not know the value of y. This is where I am confused... since the absolute value of a number is always positive, isn't it okay to divide by both sides by it and keeping the inequality sign facing the same way? While y might be positive or negative, |y| will always be positive, so why can you do this?

The simple answer is that it messes up the rest of your equation. You are correct that both |y| and y^2 will be positive, but when you divide by y, your new equation includes a y, which can now be negative. And if y is negative, x can be negative, which can't be true if the answer to this question is to be yes. Technically, if you want to solve this algebraically, you need to solve it twice, once if y is positive and once if y is negative:

if y = +
x|y| > y^2
xy > y^2
x > y

if y = -
x|y| > y^2
x(-y) > y^2 (notice I'm taking -y to take a neg y and make it pos)
-x * y > y^2
-x < y (flip the sign b/c y is neg in this scenario)
x > -y

This second scenario makes sense, b/c we don't care if y is positive or negative, but x must be (1) positive and (2) be a larger value than the absolute value of y.

[shikhers747]

Is x|y| > y^2 ?

1) x > y
2) y > 0




...


_____________
Answer = C

The explanation states that both are needed because that answer can be yes or no depending on whether x and y are positive or negative. The explanation also explicitly warns you of dividing by |y| (and eventually getting to 'Is x > y?') because you do not know the value of y. This is where I am confused... since the absolute value of a number is always positive, isn't it okay to divide by both sides by it and keeping the inequality sign facing the same way? While y might be positive or negative, |y| will always be positive, so why can you do this?

The simple answer is that it messes up the rest of your equation. You are correct that both |y| and y^2 will be positive, but when you divide by y, your new equation includes a y, which can now be negative. And if y is negative, x can be negative, which can't be true if the answer to this question is to be yes. Technically, if you want to solve this algebraically, you need to solve it twice, once if y is positive and once if y is negative:

if y = +
x|y| > y^2
xy > y^2
x > y

if y = -
x|y| > y^2
x(-y) > y^2 (notice I'm taking -y to take a neg y and make it pos)
-x * y > y^2
-x < y (flip the sign b/c y is neg in this scenario)
x > -y

This second scenario makes sense, b/c we don't care if y is positive or negative, but x must be (1) positive and (2) be a larger value than the absolute value of y.

>>>>>>>>>>>>.
Hey, joe if it is already given x>y and we can simply multiply |y| on both sides so why we need to care about positive aur negative as it is already positive...can you plz re-check its answer

[RonPurewal]

hi,
i'm sorry, but i don't understand your question.

• rather than quoting joe's entire reply—which is very, very long—please quote the portion you're asking about.

• please re-write your question so that it's easier to read. (i.e., please use punctuation, line breaks, complete sentences, etc., and use "?" at the end).

thanks.