# Math Help - Question about WHY something must be done

1. ## Question about WHY something must be done

Hi, I've been away from math for a few years now and coming back to it with an algebra review I've run into a problem.

Solve the equation $\dfrac{x}{6x-36} -7 = \dfrac{1}{x-6}$

I ended up with x = 6 , inputting that into the system gave me a "wrong answer". I noticed that plugging x=6 into the original equation causes a couple of divisions by zero so I inputted the answer as "none", that came back as correct.

Great, I got it correct! Problem is I don't really know WHY. For one I don't know why I had to plug the x=6 back into the equation to get my answer. And secondly the only reason I understand this as having "no answer" is because I saw the big scary divide by zero's come up.

I'm pretty sure I knew why I had to do these things back in high school! Too bad I can't remember now

Can anyone help me out with this? Let me know if this doesn't make any sense heh.

2. Ok, what's going on here is that in the process of solving an equation, sometimes you introduce spurious roots. Here's a very simple example: solve $x^{2}=4$ subject to $x>0.$ So the first thing you'd do, obviously, is take the square root of both sides. That gives you $x=\pm 2.$ But the problem statement said that $x$ had to be positive. Evidently, the negative solution is spurious. Therefore, the solution to the problem is exactly $x=2.$

In your case, you'd solve the problem by multiplying through by $x-6$. The problem is, $x=6$ is not in the solution space of the original problem, because you'd divide by zero. However, multiplying through by the factor $x-6$ masks that fact.

You ALWAYS have to go back and double-check your answers, to see if they fit the original problem, and if they make sense. And that is for this very reason.

Does that make sense?

3. Hmm I see. I guess the important take home message is to always check the work like you said. Thanks!

Actually I have a little bit of confusion with your post. Mainly the first part. I've observed from the quadratic formula that like you say square roots produce a negative and positive number. What I don't understand is where that comes into play with this particular problem. Was I supposed to be using square roots somewhere in the problem? I didn't >< so maybe I solved it in the wrong way.

EDIT: Actually I have a theory on why the plus/minus might be involved in this problem. After trying to solve it the way you mention (multiplying through x-6) I ended up with a quadratic equation, which I used for the quadratic forumla. Ended up with the square root being 0 so 492 plus/minus 0 divided by 81 = 6 and 6. So even though they are both positive answers (and in fact the same answer) they are still two answers. The way I originally solved the problem I actually didn't multiply through by x-6 and just rearranged and solved for X, so I guess I missed this. Is that right?

4. What I don't understand is where that comes into play with this particular problem.
I'm glad you didn't understand that, because the square root thing was only an example, intended to show you the main idea (checking your answer) by analogy. The process of solving your problem has absolutely nothing to do with square roots.

5. Originally Posted by Ackbeet
I'm glad you didn't understand that, because the square root thing was only an example, intended to show you the main idea (checking your answer) by analogy. The process of solving your problem has absolutely nothing to do with square roots.
Ah, I guess my edit must be completely off then!

So in a revision of my edit in my last post I guess the "two" answers of 6 aren't actually 2 answers because... once the zero came up in the quadratic forumla I should have just thrown it away and ended up with 1 equation of 492 / 81 ... I think! That would mean the only answer I could have is 6, at which point I plug 6 back into the original forumla to make sure it works, upon finding out that 6 does not work I'm left with no answers, which is actually the answer

Thanks for your help I think I understand this problem very well now!

$\dfrac{x}{6x-36} -7 = \dfrac{1}{x-6}.$

The first thing I would do is factor out a 6 in the denominator of the first term thus:

$\dfrac{x}{6(x-6)} -7 = \dfrac{1}{x-6}.$

Then multiply everything through by $6(x-6)$ to obtain

$x -7\cdot 6(x-6) = 6.$

That is not a quadratic, but a linear equation, right?

I didn't see your edit when I wrote post # 4. That might change your perception of post # 4 a bit.

7. Hmm I see, the way you've solved the problem is much faster than the way I went through it. I simply tried multiplying (x - 6) onto everything in the equation. After working through it I ended up with in the form 0 = 41x^2 - 492x + 1476 and used that for the quadratic forumla, which gave an answer of 6. The step I missed in the way you've done it is pulling a 6 out the denominator. Actually I'm laughing right now at how big the numbers ended up being

8. So, is everything clear now?

9. Yup, much thanks.

10. Originally Posted by cb220
Hi, I've been away from math for a few years now and coming back to it with an algebra review I've run into a problem.

Solve the equation $\dfrac{x}{6x-36} -7 = \dfrac{1}{x-6}$

I ended up with x = 6 , inputting that into the system gave me a "wrong answer". I noticed that plugging x=6 into the original equation causes a couple of divisions by zero so I inputted the answer as "none", that came back as correct.

Great, I got it correct! Problem is I don't really know WHY. For one I don't know why I had to plug the x=6 back into the equation to get my answer. And secondly the only reason I understand this as having "no answer" is because I saw the big scary divide by zero's come up.

I'm pretty sure I knew why I had to do these things back in high school! Too bad I can't remember now

Can anyone help me out with this? Let me know if this doesn't make any sense heh.
You are given the equation:

$\dfrac{x}{6x-36} -7 = \dfrac{1}{x-6}$

for this to make any sense we require that $x \ne 6$, so suppose $x \ne 6$, then:

$\dfrac{x}{6} -7(x-6) = 1$

or:

$-41x+42\times 6=6$

so:

$x=\dfrac{42 \times 6 - 6}{41}=6$

but that contradicts our assumption that $x \ne 6$ so there are no solutions.

CB

11. Originally Posted by cb220
Hi, I've been away from math for a few years now and coming back to it with an algebra review I've run into a problem.

Solve the equation $\dfrac{x}{6x-36} -7 = \dfrac{1}{x-6}$

I ended up with x = 6 , inputting that into the system gave me a "wrong answer". I noticed that plugging x=6 into the original equation causes a couple of divisions by zero so I inputted the answer as "none", that came back as correct.

Great, I got it correct! Problem is I don't really know WHY. For one I don't know why I had to plug the x=6 back into the equation to get my answer. And secondly the only reason I understand this as having "no answer" is because I saw the big scary divide by zero's come up.

I'm pretty sure I knew why I had to do these things back in high school! Too bad I can't remember now

Can anyone help me out with this? Let me know if this doesn't make any sense heh.
Another way to look at this is....

$\displaystyle\frac{x}{6x-36}-7=\frac{1}{x-6}$

which we can rewrite as

$\displaystyle\frac{x}{6(x-6)}-\frac{1}{x-6}=7$

When do the graphs of $\displaystyle\frac{x}{6(x-6)}$ and $\displaystyle\frac{1}{x-6}$ differ by 7 ?

$\displaystyle\frac{x}{6(x-6)}-\frac{6}{6(x-6)}=\frac{(x-6)}{6(x-6)}$

The graphs have a constant difference of $\displaystyle\frac{1}{6}$

while the graphs have an asymptote at x=6.

Hence the only valid equation is $\displaystyle\frac{x}{6(x-6)}-\frac{1}{6}= \frac{1}{x-6}$