Solving Equation x= (1+c+a)/(2-a): Need Help!

[koolmodee]

x=$$\frac{1+c+a(\frac{1+c+ax}{2})}{2}$$

should give x=$$\frac{1+c}{2-a}$$

I've been trying to solve it now for like two hours. Some hints?

thank you

 

[Borek]

Show what you did, it is hard to tell where is your problem. In general equation is not difficult - multiply, rearrange and you should get there.

 

[koolmodee]

x=1/2+1/2c+1/4a+1/4ac+1/4a²x

How do I get rid of the a² especially?

 

[HallsofIvy]

x=1/2+1/2c+1/4a+1/4ac+1/4a²x

How do I get rid of the a² especially?

So x- 1/4 a²x= (1- 1/4 a)x= 1/2+ 1/2 c+ 1/4 a+ 1/4 ac
Now what do you do?

 

[Borek]

So x- 1/4 a²x= (1- 1/4 a)x= 1/2+ 1/2 c+ 1/4 a+ 1/4 ac

More like

$$x-\frac 1 4 a^2x = (1- \frac 1 4 a^2)x$$

 

[koolmodee]

Dividing by a?

(1/a-a/4)x=1/2a + c/2a + 1/4 + c/4

but now?

 

[HallsofIvy]

NO! Divide by (1- 1/4 a²)!

 

[koolmodee]

Why should I do that?

Nothing cancels then on the RHS?

 

[Borek]

Divide, you will be surprised how many things cancel out

hint:

$$a^2 - b^2 = (a-b)(a+b)$$

 

[koolmodee]

Got it! thanks, guys!

 

[FordPrefect]

I am getting x = (1+c) / (1/2-a)

Is everyone else getting the "2-a" as the denominator as it was stated in the solution posted?

 

[FordPrefect]

Never-mind. I got "2-a" as my denominator now :) .

I didn't think to separate the "1" into "(2(1/2))", so that the "1/2's" could cancel out. Although, if I had not done that, and I just divided the "1/2" in the numerator, by the "1" in the denominator, to get "1/2-a", why exactly would it have been wrong?

This was the set up that I had...

x= (1/2 + 1/4a) (1+c) / (1-1/4a^2) ...

Then I rearranged the denominator so that values could cancel out...

x= (1/2 + 1/4a) (1+c) / (2(1/2)) - a(1/4a) ...where the "1/2" and the "1/4a" cancel out...

but again...what if I had changed the denominator to "1 - a(1/4a)" and got "1/2-a"?

Thanks to anyone else in advance...I hope I was clear enough with my question.

-Ford

 

[Borek]

Where did you get 1/2 from?

$$1-\frac {a^2} 4 = (1 - \frac a 2)(1 + \frac a 2)$$

It is much simpler IMHO to start by multiplying both sides of the equation by 4 at some early stage - then you end without fractions.

 

[FordPrefect]

Sorry to waste your time Borek. I noticed another error I made in my earlier calculations which screwed the whole problem up (so everyone, ignore my previous posts! haha). Since then I have fixed it and have the correct solution. Thank you anyway for your help!
-Ford

 

[snipez90]

Why all the superfluous algebraic manipulation and rearranging? Just consider the root equation satisfied here:

x = (1+c+ax)/2

Solving for x gives the desired conclusion immediately.