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
$\displaystyle x = \frac{12}{3 - \frac{3}{y}} + y$
$\displaystyle x = \frac{12y}{3y - 3} + y$
$\displaystyle x - y = \frac{12y}{3y - 3}$
$\displaystyle x - y = \frac{4y}{y - 1}$
$\displaystyle xy - y^2 - x + y = 4y$
$\displaystyle 0 = y^2 - (x-3)y + x$
$\displaystyle y = \frac{(x-3) \pm \sqrt{[-(x-3)]^2 - 4x}}{2}$
$\displaystyle x=\frac{12}{3-\frac{3}{y}}+y$
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
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
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
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!
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
Back again
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
$\displaystyle x=\frac{\frac{12}{3-\frac{3}{y}}+y}{z}$
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.