# Math Help - Tech Math(Mostly Algebra)

1. ## Tech Math(Mostly Algebra)

I'm trying to work ahead in Tech Math, and this is the only place I can find help outside of school, witch is great by the way. So I was going to just keep one thread, and add questions as I go if that's ok.

So here's the one I'm working on right now.
$C=\frac{F(R-r)}{Zt}$ Solve for Zt, F, R, & r

Can someone check these for me? I'm thinking the only one I got right was Solving Zt

Here's what I got for Zt
$Zt=\frac{(\frac{F}{R})+r}{C}$

Solve for "F"
$F=\frac{Zt(R-r)}{C}$

Solve for "R"
$R=\frac{FZt+r}{C}$

Solve for "r"
$r=\frac{ZtFR}{C}$

If someone could check this too that would be great.
$gm={(Go-G)(1+n)}$ Solve for Go

I got $Go=\frac{(1+n)+G}{gm}$

If I'm not using the forum right I'm sorry, if not could someone tell me.

2. Sorry, it looks like you haven't got any of them right. That's ok, you'll be an expert in no time. Let's do one at a time.

Solve for Zt

$
C=\frac{F(R-r)}{Zt}
$

Remember that to solve for a variable means to isolate it. So, we want to rearrange this formula so that Zt is by itself on either the right or the left side. Recall that we can manipulate the equation any way that we want, so long as we follow the Golden Rule of Algebra: Do unto one side as you do unto the other.

Step 1: Multiply both sides by Zt. This will cancel out the Zt on the right side and leave you with:

$
C\cdot Zt=F(R-r)
$

Step 2: Divide both sides by C. This cancels out the C from the left side, thereby isolating Zt:

$
Zt=\frac{F(R-r)}{C}
$

Notice that in essence, all we have done is switched C and Zt. This will always work for equations of this form.

In general, if $a = \frac {b}{c}$ then $c = \frac {b}{a}$

Do you understand this?

3. That was a great way of explaining it. I'll try to solve the rest and come back with an answer.

Well here goes an attempt. $C=\frac{F(R-r)}{Zt}$ Solve for F

$F=\frac{R-r}{ZtC}$ I'm not sure I gave it a couple of shots.

4. Solve for F

$
C=\frac{F(R-r)}{Zt}
$

We can move the Zt to the left side by multiplying both sides of the equation by it:

$
C\cdot Zt=F(R-r)
$

For this problem, consider $(R-r)$ to be one variable. Divide both sides of the equation by it to get:

$
\frac{C\cdot Zt}{(R-r)} = F
$

$
F = \frac{C\cdot Zt}{(R-r)}
$

Any questions? I'll continue answering these, if you need more clarification just let me know.

5. Originally Posted by Jarod_C
That was a great way of explaining it. I'll try to solve the rest and come back with an answer.

Well here goes an attempt. $C=\frac{F(R-r)}{Zt}$ Solve for F

$F=\frac{R-r}{ZtC}$ I'm not sure I gave it a couple of shots.
The solution for F is in my last post.

Do you see how I did it? You were close, your fraction was just reversed.

Ok, here are the steps to solve for R:

$
C=\frac{F(R-r)}{Zt}
$

1. Multiply both sides by Zt:

$
C\cdot Zt=F(R-r)
$

2. Divide both sides by F:

$
\frac {C\cdot Zt}{F} = R-r$

3. Add r to both sides:

$
\frac {C\cdot Zt}{F} + r = R$

$
R = \frac {C\cdot Zt}{F} + r$

Any questions so far on how I completed any of the steps?

6. I'm sure you're explaining this great, it looks like it. If I was awake I'm pretty sure I could follow along with your directions. I'm going to get some sleep for now, then I'll wake up tomorrow and look at this and laugh, and fell sorry for you for taking the time to help me. Seriously, Thanks. I don't know if I'm making any since right now, but that's my way of telling you I really appreciate the help.

Oh, Canada's beautiful btw I stayed by Peterborough for a week.

7. No problem if you have any questions in the morning just ask and I or someone else will get to them ASAP.

Yes, Canada is beautiful but I'm in Vancouver, opposite side of the country.