#7**+8 **

Thanks Alan,

I understand what you have presented now - I don't remember seeing it presented like that ever before :)

What I don't get is how you can use just the algebra and extract that information, for this exact question, easily.

Let me see how I would extract that:

$$\\x+a-1-xa^2=0 \qquad a\ne0,\;\;x\ne0,\;\;a+x\ne1\\\\

xa^2-1a+(1-x)=0\qquad -1=-x-(1-x)\\\\

xa^2-xa-(1-x)a+(1-x)=0\\\\

xa^2-xa+(1-x)(-a)+(1-x)=0\\\\

xa(a-1)-(1-x)(a-1)=0\\\\

(xa-1+x)(a-1)=0\\\\

$So we have the graphs of$\\\\

xa+x-1=0\qquad and \qquad a-1=0\\\\

-1+xa+x=0\qquad and \qquad a=1\\\\

a=\frac{1-x}{x}\qquad and \qquad a=1\\\\

a=\frac{1}{x}-1 \qquad and \qquad a=1\\\\$$

Now Alan, did you do all that with no working?

Maybe you use mathcad, or a scrap of paper or maybe you are just clever.

Melody
Jul 16, 2015

#2**+5 **

If you are solving for x, then x = 1/(a+1) a ≠ -1, a ≠ 0

if you are solving for a, then there are two solutions:

a = 1 x ≠ 0

and

a = -1 + 1/x x ≠ 1, x ≠ 0

.

Alan
Jul 16, 2015

#3**+5 **

IS THAT RIGHT THAT: X≠0;-a+1

A≠0;-x+1

What is the bar over the top for sabi? I will ignore it

$$a \ne 0\;\;\;\;x\ne 0\;\;\;\;x+a\ne 1$$

$$\\\frac{1}{ax}-\frac{a}{x+a-1}=0\\\\

(x+a-1)(ax)[\frac{1}{ax}-\frac{a}{x+a-1}]=0*(x+a-1)(ax)\\\\

(x+a-1)-a(ax)=0\\\\

x+a-1-a^2x=0\\\\

x-a^2x=-a+1\\\\

x(1-a^2)=1-a\\\\

x(1-a)(1+a)=1-a\\\\

1-a=0\quad or \qquad x(1+a)=1\\\\

a=1\;\;\;x\; $can be number except 0 $\quad \quad or \quad x=\frac{1}{1+a}\;\;\;\;where\;\; a\ne 1\\\\$$

remembering that

$$a \ne 0\;\;\;\;x\ne 0\;\;\;\;x+a\ne 1$$

The question that I am now asking myself is how do I determine the ends of the hole around a=1 the graphs indicate that it is more than just one point ????

---------------------------------

I will graph it but I will replace 'a' with 'y'

That is an unusual graph :)

https://www.desmos.com/calculator/ttybsu8dpw

http://www.wolframalpha.com/input/?i=1%2F%28xy%29-y%2F%28x%2By-1%29%3D0

Melody
Jul 16, 2015

#4**+5 **

*That is an unusual graph :)*

That's because it is actually *two* graphs: y = -1 + 1/x and y = 1

*The question that I am now asking myself is how do I determine the ends of the hole around a=1 the graphs indicate that it is more than just one point ????*

It's just a plotting oddity (possibly arising because of plotting two functions as though they are 1?). Plot y=-1+1/x on the same graph and you will see it overwrites the curve without a large hole around x = 1.

.

Alan
Jul 16, 2015

#5**0 **

Thanks Alan,

The bit about the plotting oddity makes sense but the bit about it being 2 graphs really does not.

I mean, I can see how it could be separated so that it is presented and plotted as two graphs.

**but** it is presented as only one graph so how can you say that it is two

Melody
Jul 16, 2015

#7**+8 **

Best Answer

Thanks Alan,

I understand what you have presented now - I don't remember seeing it presented like that ever before :)

What I don't get is how you can use just the algebra and extract that information, for this exact question, easily.

Let me see how I would extract that:

$$\\x+a-1-xa^2=0 \qquad a\ne0,\;\;x\ne0,\;\;a+x\ne1\\\\

xa^2-1a+(1-x)=0\qquad -1=-x-(1-x)\\\\

xa^2-xa-(1-x)a+(1-x)=0\\\\

xa^2-xa+(1-x)(-a)+(1-x)=0\\\\

xa(a-1)-(1-x)(a-1)=0\\\\

(xa-1+x)(a-1)=0\\\\

$So we have the graphs of$\\\\

xa+x-1=0\qquad and \qquad a-1=0\\\\

-1+xa+x=0\qquad and \qquad a=1\\\\

a=\frac{1-x}{x}\qquad and \qquad a=1\\\\

a=\frac{1}{x}-1 \qquad and \qquad a=1\\\\$$

Now Alan, did you do all that with no working?

Maybe you use mathcad, or a scrap of paper or maybe you are just clever.

Melody
Jul 16, 2015

#8**+5 **

I did it as follows:

$$\\\frac{1}{ax}-\frac{a}{x+a-1}=0\\\\1-\frac{a^2x}{x+a-1}=0\\\\x+a-1-a^2x=0\\\\\text{ or }\\\\xa^2-a+1-x=0\\\\\text{Now I just used the quadratic formula to get }\\\\a=\frac{1\pm \sqrt{1-4x(1-x)}}{2x}\\\\a=\frac{1\pm(2x-1)}{2x}\\\\a=1 \text{ and }a=\frac{1}{x}-1$$

.

Alan
Jul 16, 2015

#9**+5 **

But i've found x;exactly like melody did.is that wrong to find only x?the task is to solve the equation if its possible.so how can i know what do i need to find-x or a

sabi92
Jul 16, 2015

#10**+5 **

you have **a=1**

**and **

x as a function of a OR a as a function of x ( It makes no difference which)

I suppose since the first one has the subject a it is probably a little nicer if the second on also has a subject of a but either way is correct. That is what I think anyway. :)

So, as long as you state your restrictions, your answer is fine Sabi :)

Melody
Jul 17, 2015