# Math Help - Simplifying a simple algebra equation.

1. ## Simplifying a simple algebra equation.

Hi guys,

$A-(A-B)/(6B-5)*(6B-5)=B$

Totally suck at math (haven't done it since school and barely passed) but I need to use some simple algebra equations in some stuff I'm doing at the moment, I would appreciate any help.

My question is in the equation above I will know what number A is but I still don't know how to use it to find B.
Example when A is 16 B is 2 but I don't know how to figure it out, could you simplify this equation so that I can find B easily if I know A?

Thanks guys

2. Originally Posted by Not2l8
Hi guys,

$A-(A-B)/(6B-5)*(6B-5)=B$

Totally suck at math (haven't done it since school and barely passed) but I need to use some simple algebra equations in some stuff I'm doing at the moment, I would appreciate any help.

My question is in the equation above I will know what number A is but I still don't know how to use it to find B.
Example when A is 16 B is 2 but I don't know how to figure it out, could you simplify this equation so that I can find B easily if I know A?

Thanks guys
Hi not2l8,

Is this what you have written:

$\frac{A-(A-B)}{6B-5} \cdot (6B-5)=B$

or is it this:

$A-\frac{A-B}{6B-5} \cdot (6B-5)=B$

or something else?

3. Originally Posted by masters

$A-\frac{A-B}{6B-5} \cdot (6B-5)=B$

or something else?
Hi Masters, that is the one I was trying to write.
I can see now though that they just cancel each other out.
Mmmm... Doh, I guess that's not the equation I'm looking for.

Thanks for looking at it though

4. Originally Posted by Not2l8
Hi guys,

$A-(A-B)/(6B-5)*(6B-5)=B$
This would be

$A-\left(\frac{A-B}{6B-5}\right)(6B-5)=B$

but I am assuming that you intended the second $(6B-5)$ to be included in the denominator.

$A-\frac{A-B}{(6B-5)(6B-5)}=B$

$\Rightarrow A-\frac{A-B}{36B^2-60B+25}=B$

$\Rightarrow\frac{A\left(36B^2-60B+25\right)-(A-B)}{36B^2-60B+25}=B$

$\Rightarrow\frac{36AB^2-60AB+25A-A+B}{36B^2-60B+25}=B$

$\Rightarrow36AB^2-60AB+24A+B=36B^3-60B^2+25B$

$\Rightarrow36B^3-36AB^2-60B^2+60AB-24A+24B=0$

$\Rightarrow36B^3-(36A+60)B^2+(60A+24)B-24A=0$

Since this is a cubic in $B,$ there are at most 3 real solutions. Using the rational roots theorem, we may try $B=1,$ and using some synthetic division we do indeed find this to be a solution. The cubic then factors as

$(B-1)\left[36B^2-(36A+24)B+24A\right]=0$

Now we can simply use the quadratic formula. In the end, after factoring, we get

$(B-1)(B-A)(3B-2)=0$

Therefore, the solutions to the original equation are

$\left\{\begin{array}{rcl}
B&=&A\\
B&=&1\\
B&=&2/3\end{array}\right.$

In case it isn't clear, $B=1$ and $\mbox{B=\frac23}$ are always solutions, regardless of the value of $A.$ The third solution is $B=A.$

Edit: Okay, it seems that the problem was as you had written it. In that case, all values of $A$ and $B$ satisfy the equation.

5. Thanks very much Reckoner, I must look like such a dunce asking such simple stuff

But I 'think' I finally figured out what I'm trying to ask, I think it is this equation below -

$
(A-B)/(6B-5)=B$

Only one or two values of B make it work I think.
Like if A is 16 only B=2 makes the equation make sense.
((6B-5) is meant to the denominator in the equation above)

So how do I simplify that so it is easy to find B if I know A

6. Originally Posted by Not2l8
$
(A-B)/(6B-5)=B$

Only one or two values of B make it work I think.
Like if A is 16 only B=2 makes the equation make sense.
No, $A=16,\;B=-\frac43$ is another solution. You will never have more than two, though.

So how do I simplify that so it is easy to find B if I know A
$\frac{A-B}{6B-5}=B$

$\Rightarrow A-B=B(6B-5)$

$\Rightarrow A-B=6B^2-5B$

$\Rightarrow 6B^2-4B-A=0$

From the quadratic formula,

$B=\frac{4\pm\sqrt{16+24A}}{12}$

$=\frac{4\pm2\sqrt{4+6A}}{12}$

$=\frac13\pm\frac{\sqrt{4+6A}}6$