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
begin by subtracting y from both sides
then multiply both sides by
Multiply out the parentheses
To get the y out of the denominator, multiply both sides of the equation by y.
Set the equation equal to 0 by subtracting 12y from both sides
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
Now we divide both sides by the coefficient of the quadratic term (-3).
Now We want to build a trinomial square, but the coeficient of the linear term in y is not explicit. But...
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
We know that the left hand side factors to be
Taking the root
Subtracting
Simplifying
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.
.
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:
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
Multiply both sides by z
Multiply Numerator and denominator by y
simplify the fraction
Multiply both sides by the denominator
simplify
Move every thing to one side
factor
now we take note that this is of the form
so now we just plug "a,b, andc" where they're supposed to go in the formula
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.