# Math Help - I need to know what this is called.

1. ## I need to know what this is called.

Hi, guys. We are learning about an an algebraic proportional equality (this is not what it is called, just a description of sorts) something in class right now, and I haven't been able to comprehend it. I asked my professor what it was called and he said it didn't really have a name and that it was just an application of different algebraic principles on ratios. So I don't know what to look for specifically online to discover comprehension on my own. My best idea thus far is to post an example here and see if you guys can point me in the right direction as to how to learn it. It just hasn't clicked with me yet, as it is a different way to think about algebra.

So, here is an example (well, I will post a couple, one considered hard and one considered easy):

Given,
$\theta = \theta_\circ + \omega_\circ t + \frac{1}{2}\alpha t^2$

if you $\theta$ to $2\theta$ what would you to to $\alpha$ to make both sides equal?

And another:

Given,

$\frac{36 + xy}{uw} = z$

what would you do to $x$ to make both sides equal if we changed $z$ to $3z$? What about $w$?

I'm not looking for the answers here. I'm looking to understand. I know that you can't change the variables themselves (as in giving it an exponent), you have to add an additional coefficient beside them.

Any help?

2. (36 + xy) / (uw) = z

If I follow what you're saying, each side of the equation is multiplied by 3
(giving value 3z to right side), but only the x is affected on the left side.

So we'll let x's new value = k:
(36 + ky) / (uw) = 3z
36 + ky = 3uwz
ky = 3uwz - 36
k = (3uwz - 36) / y

So the value of x needs to be changed to the equivalent of (3uwz - 36) / y

Best I can do for you; 1st time I see such a problem.

3. Thanks! I want to say that that is exactly how it is done, but I don't have the answers. Still, I believe you are right, and I understand exactly what you did. I resembles what he was doing on the board but in a way I can grasp, and I honestly don't see another way to approach it. Thanks again! I will try working the problems using this method. If I get them wrong, it will still help me learn it by understanding my errors.