For which values of x and y will k have a positive value?

[Hancko]

To whom it may concern,

I am doing a preparatory course for mathematics. I stumbled across this question, but I cannot seem to figure it out. The answer is: x=2; y=1.
Could someone please explain to me how to get to this solution? Thank you.

 

[Deleted member 4993]

To whom it may concern,

I am doing a preparatory course for mathematics. I stumbled across this question, but I cannot seem to figure it out. The answer is: x=2; y=1.
Could someone please explain to me how to get to this solution? Thank you.
View attachment 25177

Simply evaluate "k" at the given values of 'x' & 'y' (by hand or by calculator or mentally) and decide.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Cubist]

A shortcut is available if you're doing it in your head. Will the following give a positive or negative result?

[math] \frac{-}{+}\times\frac{+}{-}[/math]

 

[Dr.Peterson]

I am doing a preparatory course for mathematics. I stumbled across this question, but I cannot seem to figure it out. The answer is: x=2; y=1.
Could someone please explain to me how to get to this solution? Thank you.
View attachment 25177

This is not asking you to actually solve anything (to find all pairs for which k is positive -- though that can be done), but just to check each of the four pairs. I don't care for that sort of misleading problem, where the fact that it is multiple-choice totally changes the genre of the problem, and makes it harder if you use higher-level skills than if you "cheat".

To see whether they are right, just replace x with 2 and y with 1:

[MATH]\frac{4(2)^2-9(1)^2}{2(2)+(1)}\times\frac{2(2)^2-(2)(1)-(1)^2}{2(2)-3(1)}=\frac{7}{5}\times\frac{5}{1}=7[/MATH]​

Looks good. But to be sure, you need to check the rest.

Now, if you want to practice solving rational inequalities, just simplify the expression for k and then divide numerator and denominator by x to get an expression in y/x. Then you can solve the inequality and find interval(s) in which the result is positive. But that is definitely overkill (though just doing the simplification might save some work).

 

[Hancko]

This is not asking you to actually solve anything (to find all pairs for which k is positive -- though that can be done), but just to check each of the four pairs. I don't care for that sort of misleading problem, where the fact that it is multiple-choice totally changes the genre of the problem, and makes it harder if you use higher-level skills than if you "cheat".

To see whether they are right, just replace x with 2 and y with 1:

[MATH]\frac{4(2)^2-9(1)^2}{2(2)+(1)}\times\frac{2(2)^2-(2)(1)-(1)^2}{2(2)-3(1)}=\frac{7}{5}\times\frac{5}{1}=7[/MATH]​

Looks good. But to be sure, you need to check the rest.

Now, if you want to practice solving rational inequalities, just simplify the expression for k and then divide numerator and denominator by x to get an expression in y/x. Then you can solve the inequality and find interval(s) in which the result is positive. But that is definitely overkill (though just doing the simplification might save some work).

Dear Peterson,
Thank you for your response. I have used your method and have arrived at the correct answer by replacing x with 2 and y with 1. However, I would like to know will I be asked this question without it being multiple choice in the future? How do I find the solutions from the intervals? The intervals I got are:

2x+y>0 OR x-y>0

 

[Hancko]

A shortcut is available if you're doing it in your head. Will the following give a positive or negative result?

[math] \frac{-}{+}\times\frac{+}{-}[/math]

I have done so and this will give a positive result. I do not, however, understand how to get to the solution with the intervals.

 

[Hancko]

Simply evaluate "k" at the given values of 'x' & 'y' (by hand or by calculator or mentally) and decide.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

Hello, I have evaluated and got the correct answer, however, I want to know how I would go about solving this problem without it being multiple choice.

I have gone further and have ended up with these two intervals:
2x+3y>0 OR x-y>0

How do I get to the solution of x=2 and y=1 with these intervals?

Thank you for your support.

 

[Deleted member 4993]

To whom it may concern,

I am doing a preparatory course for mathematics. I stumbled across this question, but I cannot seem to figure it out. The answer is: x=2; y=1.
Could someone please explain to me how to get to this solution? Thank you.
View attachment 25177

\(\displaystyle k \ = \ \frac{4x^2-9y^2}{2x + y} \ * \frac{2x^2 - xy - y^2}{2x - 3y}\)

\(\displaystyle k \ = \ \frac{(2x+3y)(2x-3y)}{2x + y} \ * \frac{(2x+y)(x-y)}{2x - 3y}\)

k = (2x + 3y) * (x - y) ........ continue

 

[Dr.Peterson]

Dear Peterson,
Thank you for your response. I have used your method and have arrived at the correct answer by replacing x with 2 and y with 1. However, I would like to know will I be asked this question without it being multiple choice in the future? How do I find the solutions from the intervals? The intervals I got are:

2x+y>0 OR x-y>0

I can't tell what you will be expected to do in the future based on this problem, because, as I said, it is misleading. You need to look at your syllabus, or ask.

As SK showed how to simplify the expression, you would end up solving the quadratic inequality (2x + 3y)(x - y) > 0. Your conclusion, that one of the two factors must be positive, is incorrect; you can get a positive product either as +*+ or as -*-. So your next step is to solve this:

Either (2x + 3y) > 0 and (x - y) > 0, or (2x + 3y) < 0 and (x - y) < 0​

Each case will lead to a region on the plane, and you can then take the union.

(By the way, I previously made a suggestion that was based on doing part of the work in my head, and visualizing it wrong; I saw it as (2x + 3y)/(x - y) > 0, which has some very different implications.)

 

[Hancko]

\(\displaystyle k \ = \ \frac{4x^2-9y^2}{2x + y} \ * \frac{2x^2 - xy - y^2}{2x - 3y}\)

\(\displaystyle k \ = \ \frac{(2x+3y)(2x-3y)}{2x + y} \ * \frac{(2x+y)(x-y)}{2x - 3y}\)

k = (2x + 3y) * (x - y) ........ continue

Thank you for your support.

 

[Hancko]

I can't tell what you will be expected to do in the future based on this problem, because, as I said, it is misleading. You need to look at your syllabus, or ask.

As SK showed how to simplify the expression, you would end up solving the quadratic inequality (2x + 3y)(x - y) > 0. Your conclusion, that one of the two factors must be positive, is incorrect; you can get a positive product either as +*+ or as -*-. So your next step is to solve this:

Either (2x + 3y) > 0 and (x - y) > 0, or (2x + 3y) < 0 and (x - y) < 0​

Each case will lead to a region on the plane, and you can then take the union.

(By the way, I previously made a suggestion that was based on doing part of the work in my head, and visualizing it wrong; I saw it as (2x + 3y)/(x - y) > 0, which has some very different implications.)

Thank you for your support.