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IS THAT RIGHT THAT: X≠0;-a+1

                                  A≠0;-x+1

sabi92  Jul 15, 2015

Best Answer 

 #7
avatar+91001 
+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
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9+0 Answers

 #2
avatar+26328 
+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
avatar+91001 
+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
avatar+26328 
+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
avatar+91001 
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
 #6
avatar+26328 
+5

It plots separate solutions from a single equation in a single colour. e.g. 

 Desmos multiple solutions graph

.

Alan  Jul 16, 2015
 #7
avatar+91001 
+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
avatar+26328 
+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
avatar+262 
+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
avatar+91001 
+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

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