Solving for y: 2 - (4/y) = (16/y^2)

[snb781]

I know this question is easy, but I've tried several approaches already with no luck, and I just can't seem to figure it out or even remember what this type of problem would be called. It's so easy I can tell just by looking at it that the answer is 4, but I can't remeber how to solve this type of problem, and my textbook is of no help.

2 - (4/y) = (16/y^2)

I would greatly appreciate any hints or recommendations about where to find information. Thank you.

 

[JakeD]

Re: Solving for y

I know this question is easy, but I've tried several approaches already with no luck, and I just can't seem to figure it out or even remember what this type of problem would be called. It's so easy I can tell just by looking at it that the answer is 4, but I can't remeber how to solve this type of problem, and my textbook is of no help.

2 - (4/y) = (16/y^2)

I would greatly appreciate any hints or recommendations about where to find information. Thank you.

Hi, multiply both sides of the equation by y^2 and then use the quadratic formula.

 

[snb781]

I knew it was something obvious. Actually, I don't even have to do that, but you turned me in the right direction, so thank you very much.

I could just do this:

2 - (4/y) = (16/y^2)

2 = (16/y^2) + (4/y)

0 = (16/y^2) + (4/y) -2

and then plug that into the quadratic equation (a=1/16, b=1/4, c=-2) (because (16/y^2) is the same as (1/16)(y^2), for example), the result of which is 4 and -8, with -8 being an extraneous solution.

Thanks again.

 

[JakeD]

I knew it was something obvious. Actually, I don't even have to do that, but you turned me in the right direction, so thank you very much.

I could just do this:

2 - (4/y) = (16/y^2)

2 = (16/y^2) + (4/y)

0 = (16/y^2) + (4/y) -2

and then plug that into the quadratic equation (a=1/16, b=1/4, c=-2) (because (16/y^2) is the same as (1/16)(y^2), for example), the result of which is 4 and -8, with -8 being an extraneous solution.

Thanks again.

Huh? (16/y^2) is not the same as (1/16)(y^2). Solving it as I said yields the solutions 4 and -2 to the original equation. -8 is not a solution to the original equation.

What is "extraneous" about the negative solution? -2 is a valid solution to the original equation, although it may be ruled out by other restrictions that you've haven't shown here.