Associated Topics || Dr. Math Home || Search Dr. Math

### Isolating a Variable in a Tricky Place, or Raised to a Power

```Date: 10/01/2010 at 16:50:25
From: Trevor
Subject: Isolating Variables

Well, I am trying to solve for specific variables in equations like this:

x = y + v(t) + 0.5(a)(t)^2

I can solve this equation for several of the variables -- but there are
multiple t's to isolate, and a is in a very complicated place.

I've tried to isolate t by moving y and 0.5(a)(t)^2 to the opposite side
via subtraction. But that left me with terms in t on both sides of the
equality.

I tried moving certain terms back to the other side of the problem to
isolate t, but even then I had more than one t. It seemed like a Rubik's
Cube, the way I would move one part only to mess up another.

I would love some help on how to do this. Thanks!

```

```
Date: 10/03/2010 at 10:57:39
From: Doctor Ian
Subject: Re: Isolating Variables

Hi Trevor,

To isolate a, you would isolate 0.5(a)t^2, and then divide both sides by
(0.5t^2). So that should be straightforward.

Substitution can be very handy when a variable "is in a very complicated
place." That is, you can write

x = y + v(t) + 0.5(a)(t)^2

as

this = that + other*a           this  = x
that  = y + v(t)
other = 0.5*t^2

The only restriction here is that a shouldn't appear in any of the
expressions for which you're substituting.

Once you have that, isolating a is pretty easy:

this - that = other*a

this - that
----------- = a
other

Then you can undo the substitutions:

[x] - [y + v(t)]
---------------- = a
[0.5*t^2]

I like to wrap everything in ()'s or []'s to avoid any possible confusion

But as far as isolating 'multiple t's,' the difficulty is not so much that
that variable appears twice. For example, your equation has just as many
t's as this one does ...

x = at + bt

... and you could readily combine those terms:

x = (a + b)t

x/(a + b) = t

The difficulty is that this isn't a linear equation, so you really are
looking for a different kind of solution -- not one where you isolate the
variable, but one where you find a product of binomials, e.g.,

0 = t^2 + 5t + 6

= (t + 2)(t + 3)          True when t = -2 or -3

Do you recognize x = y + v(t) + 0.5(a)(t)^2 as a quadratic equation in t?
In the general case, there are no formulas for isolating a variable in a
polynomial -- but there are for quartics, cubics, or quadratics such as
yours. Isolating t in this context is the same as using the quadratic
formula. You can just apply that directly, or you can look in our archives
to see how the formula is derived, and apply those same steps one by one.

Does it all make more sense now?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 10/14/2010 at 16:18:12
From: Trevor
Subject: Thank you (Isolating Variables)

Thank you so much, that is exactly what I needed to know.

You explained it perfectly and I actually understand it now.

Thanks!
```
Associated Topics:
High School Polynomials

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search