Algebra Question

[Sariani]

Here's a little question that I've been tinkering with and according to my answer book, the answer should be "Empty Set" but for some blasted reason I keep getting y = -5.

I put the number into the equation and found that I'll end up dividing my Zero, which can't be done ... but I'm not certain if that's 'how' I'm supposed to know that the set is empty or if I just fudged my numbers. My math book only seems to go over nice and clean questions, but the question sets get fuzzy!



Sorry for the bad writing, I'm not good with a mouse. sweatdrop  

[Sioga]

Does the question say anything about "extraneous solutions"? Have you learned of these kinds of solutions? I solved the problem you gave two different ways and I got y = -5 both times as well. And when I plugged it back in, it didn't work because it gave 0 as two of the three denominators. What the book might be giving as "Empty Set" means that there is no real answer to the problem if you discount the extraneous solution of y = -5.

But that's just me. I just finished Algebra 2 in school and I had one or two of these kinds of problems on the exam. And I got a really good grade, so I think that means I know what I'm doing. But I may be wrong. What you have up there isn't incorrect, technically, but they might be looking for a different solving method.

 

[Sariani]

It doesn't say anything about extraneous solutions, the question is just what is written above. I'm usually pretty good with my math ... but I'm accustomed to empty sets essentially defining themselves in the solution as opposed to having to be actually plugged in and double checked!


Usually I'll get an answer like:

23 = 35

or

2 + y = y

but this is the first time that I got a defined variable that was still an empty set! It's just ... really weird. eek

BTW, Thank you for replying!  

[VorpalNeko]

You did it correctly up to the y = -5. As Sioga says, that solution is extraneous. You'll probably have lots more cases like that. When you have polynomials in the denominator, square roots, or logarithms, etc., it's not uncommon to look like you're getting solutions at first and have to discard some or even all of them.

Easiest way to think of it is to just check: does it work in the original equation? And this doesn't. Byt what's really going on is when you're multiplying through by (y²-25) when you got rid of the denominator, you're masking the potential division by zero when y is -5 or +5. In other words, you're implicitly assuming that it doesn't happen, i.e., y is neither of those numbers.

So really the last steps should have a condition attached:
-y + 5 - 9y - 45 = 10,        as long as y ≠ ±5
-10y = 50,                    y ≠ ±5
y = -5,                     y ≠ ±5
Most of the time, people generally don't bother to write it, but it's still there. So the solution literally reads "y = -5, but y is not +5 or -5". And that's just as much a contradiction as "2 + y = y" ("2 = 0").  

[Sioga]

I remember that! The whole, y = 5 for y ≠ ±5 thing, that is. I learned that a couple of months into the second semester of this past school year. Good point, that most people leave that out. My textbook called them restrictions.