Understanding Quotients in Equations

[Richie Smash]

<Moderator's note: Moved from a technical forum and thus no template.>

Can someone explain this to me?

There's an equation D2V2 = 4D1V1.

I understand how to get V2 = (4D1V1)/D2.
And I also know how to get V1 = (D2V2)/(4D1)

From that equation I had up there.

Now say V1/V2 = D2/(4D1)

And A = 5D1V1
B = V1 + V2

Say if I want to divide both sides by V2
I know that B/V2 will give me 1+ V1/V2

But I do not know at all, I can't grasp the concept, how A/V2 = 5D1(V1/V2).

Can someone explain, please don't be hostile, I'm simply trying to make myself better in this life by furthering my education :)

 

[Bandersnatch]

Just for clarification, is you question: how to get from
$$\frac{A}{V_2}=\frac{5D_1V_1}{V_2}$$
to
$$\frac{A}{V_2}=5D_1 \frac{V_1}{V_2}$$
?

 

[kuruman]

It is true that
$$\frac{(5D1*V1)}{V2}=5D1*\left( \frac{V1}{V2}\right) =\left( \frac{5D1}{V2}\right) *V1$$
where the parentheses indicate which quantity you calculate first. In other words it doesn't matter if you first find the product 5D1*V1 and then divide by V2 or find one of the ratios first and then multiply the ratio by the quantity you didn't use.
Example
Say 5D1 = 6, V1 = 4 and C2 = 2. Then
$$\frac{(5D1*V1)}{V2}=\frac{(6*4)}{2}=\frac{24}{2}=12.$$ $$5D1*\left( \frac{V1}{V2}\right)=6*\left( \frac{4}{2}\right)=6*2=12.$$ $$\left( \frac{5D1}{V2}\right) *V1=\left( \frac{6}{2}\right) *4=3*4=12.$$

 

[Richie Smash]

Just for clarification, is you question: how to get from
$$\frac{A}{V_2}=\frac{5D_1V_1}{V_2}$$
to
$$\frac{A}{V_2}=5D_1 \frac{V_1}{V_2}$$
?

Yes this was my question.

 

[Richie Smash]

It is true that
$$\frac{(5D1*V1)}{V2}=5D1*\left( \frac{V1}{V2}\right) =\left( \frac{5D1}{V2}\right) *V1$$
where the parentheses indicate which quantity you calculate first. In other words it doesn't matter if you first find the product 5D1*V1 and then divide by V2 or find one of the ratios first and then multiply the ratio by the quantity you didn't use.
Example
Say 5D1 = 6, V1 = 4 and C2 = 2. Then
$$\frac{(5D1*V1)}{V2}=\frac{(6*4)}{2}=\frac{24}{2}=12.$$ $$5D1*\left( \frac{V1}{V2}\right)=6*\left( \frac{4}{2}\right)=6*2=12.$$ $$\left( \frac{5D1}{V2}\right) *V1=\left( \frac{6}{2}\right) *4=3*4=12.$$

Hey thank you very much Kuruman that really did clear things up!

 

[Richie Smash]

Hi guys, Now that I understand the concept above,

I've been trying and trying to get a value for D1 from this equation

D3 = 5D1*(V1/V2)/(1 +(V1/V2))
I want to make D1 the subject.

I know the value for V1/V2 is D2/(4D1),

I've literally been trying over and over and over but Keep ending up with complicated or the wrong answer each time.

 

[kuruman]

First move (V1/V2) over to the other side and change its sign.
Then divide both sides by 5*V1/V2.

 

[Richie Smash]

I got D1 = D3*(1+V1/V2)

 

[Mark44]

D3 = 5D1*(V1/V2)/1 +(V1/V2)
I want to make D1 the subject.

You have used lots of parentheses, but probably not as many as you need.
The right side of the equation above should probably be written $5D_1 * (V_1/V_2)/(1 + (V_1/V_2))$
Without that extra pair of parentheses, the expression is $5D_1 * \frac{\frac{V_1}{V_2}}{1} + \frac{V_1}{V_2}$. I'm fairly sure this isn't what you meant.

First move (V1/V2) over to the other side and change its sign.
Then divide both sides by 5*V1/V2.

I don't think so. See above.

 

[Mark44]

I got D1 = D3*(1+V1/V2)

No. Where did the 5 go?

 

[Richie Smash]

Yes you were right I needed extra parentheses

Ok I tried again and Got
I got D1 = (D3*(1+V1/V2))/(5*V1/V2)

 

[kuruman]

I don't think so. See above.

I didn't think so either, but I had to respond to what was posted by OP.

 

[Richie Smash]

Yes you were right I needed extra parentheses

Ok I tried again and Got
I got D1 = (D3*(1+V1/V2))/(5*V1/V2)

Is this correct?

 

[kuruman]

Is this correct?

Looks right to me.

 

[Mark44]

Ok I tried again and Got
I got D1 = (D3*(1+V1/V2))/(5*V1/V2)

It's correct, but can be simplified. Dividing by V1/V2 is the same as multiplying by its reciprocal; that is, multiplying by V2/V1.
D1 = (D3*(1+V1/V2))/(5*V1/V2) = $\frac{D_3}5 (1 + \frac{V_1}{V_2})\frac{V_2}{V_1} = \frac{D_3}5 (\frac{V_2}{V_1} + 1)$

 

[Richie Smash]

Ok Guys, and here is the final piece of the puzzle :D

I have values for D2 and D3, which are D3 = 8.56 and D2 = 8.9

Now with the equation that we have simplified, I also have a value for V2/V1

Which is in terms of D1 and D2 and that value is (4D1)/D2
Now we can substitute that value back into our equation.

So D1 = D3/5 *(V2/V1 +1)

This will become D1 = D3/5*((4D1)/D2 +1)

Now as You can see, I still have a 4D1 on the right side, but I don't want that, I want all the D1s over on the left side, and ultimately I can solve for D1 using my two real values for D3 and D2, but I'm now unsure as to how to do that.

I'm currently self taught, and I learn as I go.

 

[kuruman]

I also have a value for V2/V2

You mean V1/V2, of course.

...but I'm now unsure as to how to do that.

To isolate D1, the first step is to get rid of the parentheses and distribute the term D3/5 by multiplying both terms between them by it.

 

[Richie Smash]

I think I can divide both side by D3/5 and I will get

(5D1)/D3 = ((4D1)/D2 +1)

 

[kuruman]

That was my first thought and also works as a first step. Now what do you propose to do?

 

[Richie Smash]

I was thinking dividing both sides by 5D1, but It's proving to be confusing

 

[kuruman]

You have D1 on both sides of the equation. The goal is to have D1 on only one side of the equation. So ... ?

On edit: Can one of them be moved?

 

[Richie Smash]

Yes the left side can be divided by D1 I believe, and I will be left with 5/D3 = 4/D2+1/D1

 

[kuruman]

OK. Now what do you need to do to have D1 alone on one side?

 

[Richie Smash]

My best guess is to subtract 4/D2 from both sides.

 

[kuruman]

Go for it. Let me give you a hint about this business. After you divide both sides of an equation by the same quantity, a very good guess for the next step is to add or subtract something from both sides. Dividing both sides by some other quantity will not buy you much because you have already done so.

 

[Richie Smash]

Ahh, thank you Kuruman... I have solved the beast... I feel so relieved.

1/D1= 5/D3-4/D2
1/D1 = 5/8.56 - 4/8.9
1/D1 = (44.5-34.24)/76.184
1/D1 = 0.135 to 3SF
D1 = 1/0.135
D1 = 7.41 to 3 Sf :D

 

[Mark44]

Ahh, thank you Kuruman... I have solved the beast... I feel so relieved.

1/D1= 5/D3-4/D2
1/D1 = 5/8.56 - 4/8.9
1/D1 = (44.5-34.24)/76.184
1/D1 = 0.135 to 3SF

What does the "to 3SF" mean?

D1 = 1/0.135
D1 = 7.41 to 3 Sf :D

Same here.

 

[Richie Smash]

What does the "to 3SF" mean?
Same here.

It means to three significant figures.

 

[Mark44]

It means to three significant figures.

That's not a commonly known acronym.

Also, if you round early on, as you did here -- 1/D1 = 0.135 to 3SF --
you are likely to lose precision in later calculations. In other words, don't round until the very end.