Hi,

I have the following formula which i need expressed in terms of Y.

X = 12/(3-(3/Y)) + Y

Any help would be appreciated

Doug

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- Jun 19th 2009, 05:32 PMboohahMake Y the subject of the formula
Hi,

I have the following formula which i need expressed in terms of Y.

X = 12/(3-(3/Y)) + Y

Any help would be appreciated

Doug - Jun 19th 2009, 05:43 PMskeeter
$\displaystyle x = \frac{12}{3 - \frac{3}{y}} + y$

$\displaystyle x = \frac{12y}{3y - 3} + y$

$\displaystyle x - y = \frac{12y}{3y - 3}$

$\displaystyle x - y = \frac{4y}{y - 1}$

$\displaystyle xy - y^2 - x + y = 4y$

$\displaystyle 0 = y^2 - (x-3)y + x$

$\displaystyle y = \frac{(x-3) \pm \sqrt{[-(x-3)]^2 - 4x}}{2}$ - Jun 19th 2009, 05:57 PMStroodle
I get:

$\displaystyle y=\frac{(x-3)\pm\sqrt{x^2-10x+9}}{2}$ - Jun 19th 2009, 06:57 PMboohah
I am with you stroodle, the values I am getting with your formula match the theoretical values I should be getting exactly. Many thanks to you and Skeeter for your help, much appreciated.

Doug - Jun 19th 2009, 08:05 PMVonNemo19
- Jun 20th 2009, 02:19 AMboohah
Hey Stroodle

Any chance of you showing the steps you used to come up with your answer.

Doug - Jun 20th 2009, 09:40 AMVonNemo19

$\displaystyle x=\frac{12}{3-\frac{3}{y}}+y$

begin by subtracting y from both sides

$\displaystyle x-y=\frac{12}{3-\frac{3}{y}}$

then multiply both sides by $\displaystyle \left(3-\frac{3}{y}\right)$

$\displaystyle \left(3-\frac{3}{y}\right)(x-y)=\frac{12}{3-\frac{3}{y}}\left(3-\frac{3}{y}\right)$

Multiply out the parentheses

$\displaystyle 3x-3y-\frac{3}{y}x+3=12$

To get the y out of the denominator, multiply both sides of the equation by y.

$\displaystyle y\left(3x-3y-\frac{3}{y}x+3\right)=(12)y$

$\displaystyle 3xy-3y^2-3x+3y=12y$

Set the equation equal to 0 by subtracting 12y from both sides

$\displaystyle 3xy-3y^2-3x-9y=0$

Notuice that we have a quadratic in y. In order to solve for y in terms of x, we need to employ a little ingenuity.

We have to do this.

We keep terms with y on the left

$\displaystyle 3xy-3y^2-9y=3x$

Now we divide both sides by the coefficient of the quadratic term (-3).

$\displaystyle -xy+y^2+3y=-x$

$\displaystyle y^2+3y-xy=-x$

Now We want to build a trinomial square, but the coeficient of the linear term in y is not explicit. But...

$\displaystyle y^2+(3-x)y=-x$

And we see that the coefficient is (3-x).

So to build a trinomial square, we take half of the coefficient of y, squatre it, and add it to both sides

$\displaystyle y^2+(3-x)y+\left[\frac{(3-x)}{2}\right]^2=-x+\left[\frac{(3-x)}{2}\right]^2$

We know that the left hand side factors to be

$\displaystyle \left(y+\frac{3-x}{2}\right)^2=-x+\left[\frac{(3-x)}{2}\right]^2$

Taking the root

$\displaystyle y+\frac{3-x}{2}=\pm\sqrt{-x+\left[\frac{(3-x)}{2}\right]^2}$

Subtracting

$\displaystyle y=\pm\sqrt{-x+\left[\frac{(3-x)}{2}\right]^2}-\frac{3-x}{2}=$

Simplifying

$\displaystyle y=\pm\frac{\sqrt{(3-x)^2-4x}}{2}-\frac{3-x}{2}$

$\displaystyle =\frac{-(3-x)\pm\sqrt{(x-3)^2-4x}}{2}$

Notice the negative - Jun 20th 2009, 01:46 PMboohah
Unreal,

I really appreciate the detail you took in explaining it VonNemo, me being very rusty at this sort of thing. Thank you heaps for all your help.

Doug - Jun 20th 2009, 03:45 PMVonNemo19
If you're rusty check this out:

http://www.mathhelpforum.com/math-he...c-formula.html

You'll never have trouble with quadratics again.

Hopefully(Wink) - Jun 20th 2009, 04:45 PMboohah
Hi VonNemo

I have checked out your page on quadratic equations and it is going to become my bible over the following weeks. The reason I asked to have that formula solved is because i have many similar formula that I need to draw up for a computer simulation. Although the values vary, the basic structure of the formulae look like the one i provided. There are a couple though that introduce another variable and I was wondering if the same method could be used to solve those.

X = (12/(3-(3/Y))+Y)/Z

What techniques would i need to solve the same equation now that a third term has been added.

Doug - Jun 20th 2009, 05:18 PMVonNemo19
If you are trying to solve for y here in terms of x and z, then treat everything except for y as a constant. Of course x and z are variables, but for the sake of argument, pretend for the time being that they are not.

You want something that resembels the quadratic equation in standard form, i.e.

$\displaystyle ax^2+bx+c=0$ .

Remember that a, b, and c, can themselves all be very complicated expressions, but they are still just a, b, and c in terms of the quadratic formula.

Here's an example of what I mean:

Look at this:

$\displaystyle \underbrace{(z^2+2xz)}_ay^2+\underbrace{(\frac{z^2 +xz+x^2}{4x})}_by+\underbrace{(x-z)}_c=0$

Do you see what's happening?

Now all you would have to do to solve for y is substitute a, b, and c into their rightful place in the quadratic formula!

Peace-Out!(Rock) - Jun 20th 2009, 05:31 PMboohah
Hi VonNemo

I was working it out using the solutions you used in your previous posts and I had just came to the same conclusion, was working it out on paper and I logged back on to say I had figured it out and not to worry. All of which is great because your solutions have given me the tools I am going to need to get all of this done.

I can not thank you enough for all of your effort, thanks mate, you have taken a world of weight off my shoulders.

Doug - Jun 20th 2009, 05:40 PMVonNemo19
- Jun 20th 2009, 08:13 PMboohah
Back again (Smirk)

I am having one more problem, the solution i get to the formula with the extra variable in it is as follows:

$\displaystyle

y = \frac{-(3-zx) \pm \sqrt{(3-zx)^2 - 4zx}}{2}

$

Unfortunately that isnt producing the results I am looking for and I fear I have made a mistake. What do you think.

Doug - Jun 20th 2009, 08:55 PMVonNemo19
$\displaystyle x=\frac{\frac{12}{3-\frac{3}{y}}+y}{z}$

Multiply both sides by z

$\displaystyle xz=\frac{12}{3-\frac{3}{y}}+y$

Multiply Numerator and denominator by y

$\displaystyle xz=\frac{12}{3-\frac{3}{y}}\cdot\frac{y}{y}+y$

$\displaystyle xz=\frac{12y}{3y-3}+y$

simplify the fraction

$\displaystyle xz=\frac{4y}{y-1}+y$

Multiply both sides by the denominator

$\displaystyle (y-1)xz=\frac{4y}{y-1}(y-1)+y(y-1)$

simplify

$\displaystyle xzy-xz=4y+y^2-y$

$\displaystyle xzy-xz=y^2+3y$

Move every thing to one side

$\displaystyle 0=y^2+3y-xzy+xz$

factor

$\displaystyle 0=y^2+(3-xz)y+xz$

now we take note that this is of the form $\displaystyle ay^2+by+c=0$

so now we just plug "a,b, andc" where they're supposed to go in the formula

$\displaystyle y=\frac{-(3-xz)\pm\sqrt{(3-xz)^2-4(1)(xz)}}{2(1)}$

you can see that you are correct. Perhaps you are supposed to simply solve for z. In functions of three variables, z is usually written as a function of x and y for the sake of convention.