This is like a fifth grade question

[flyingpig]

Homework Statement

Let's say I have $$V = P\frac{a}{b+a}$$

Solve for a

So I did the math and I got $$\frac{Vb}{V+P}$$

Then I decided to do it again and tried to invert both sides of $$V = P\frac{a}{b+a}$$ to $$V^-1 = P^-1(\frac{a}{b+a})^-1$$

Now my question is, why isn't $$(\frac{a}{b+a})^-1$$ = $$\frac{b+a}{a}$$

 

[Pengwuino]

For one, your 2nd line is incorrect, check your math.

Second, why should they be equal?

 

[Mark44]

Homework Statement

Let's say I have $$V = P\frac{a}{b+a}$$

Solve for a

So I did the math and I got $$\frac{Vb}{V+P}$$

For one thing, you should have ended with an equation. More importantly, your work is incorrect.

If you multiply your first equation by b + a, you get
Vb + Va = Pa
==> Va - Pa = -Vb
Now, factor a out of the two terms on the left and solve for a. You don't get a = Vb/(V + P), which is what I think you were trying to say above.

Then I decided to do it again and tried to invert both sides of $$V = P\frac{a}{b+a}$$ to $$V^-1 = P^-1(\frac{a}{b+a})^-1$$

Now my question is, why isn't $$(\frac{a}{b+a})^-1$$ = $$\frac{b+a}{a}$$

It is, but that doesn't do you much good if you're trying to solve for a.

 

[flyingpig]

You know what I don't know why I changed b+a to b-a on my paper lol

The real correct answer should have been

Vb/(P - V) = a

Secondly, are you guys saying it is equal to that?

 

[Mentallic]

Yes, $$\left(\frac{a+b}{a}\right)^{-1}=\frac{a}{a+b}$$ as you stated.

 

[zketrouble]

Homework Statement

Let's say I have $$V = P\frac{a}{b+a}$$

Solve for a

So I did the math and I got $$\frac{Vb}{V+P}$$

Then I decided to do it again and tried to invert both sides of $$V = P\frac{a}{b+a}$$ to $$V^-1 = P^-1(\frac{a}{b+a})^-1$$

Now my question is, why isn't $$(\frac{a}{b+a})^-1$$ = $$\frac{b+a}{a}$$

Forgive me for not being familiar with the whole math print thing.

So you have V = P[a/(a+b)]
a does not = vb/(V+p), not sure where you got this from.

V= P*[a/(a+b)]

I would start by multiplying both sides of the equation by (a+b) to get the fraction out of the original equation and then worry about solving for a. How did you get the A = vb/(V+p) figure?

 

[flyingpig]

If $$(\frac{a}{b+a})^-1$$ = $$\frac{b+a}{a}$$ indeed, then couldn't I have done this

$$V^-1 = P^-1(\frac{a}{b+a})^-1$$

$$\frac{P}{V} = \frac{b+a}{a}$$$$\frac{P}{V} = \frac{b}{a} + \frac{a}{a}$$
$$\frac{P}{V} = \frac{b}{a} + 1$$

$$\frac{P}{V} = \frac{b}{a} + 1$$

$$\frac{P}{V} - 1 = \frac{b}{a}$$

$$a = \frac{b}{]\frac{P}{V} - 1}$$

Clearly this is wrong

 

[madah12]

a=Vb/(P - V)
divide that by v in the numerator and denominator and you will get what you wrote
a= (vb/v) / ((p -v)/v)
=b/(p/v -1)

 

[Char. Limit]

Aside from that, since when is this fifth grade? When I was in fifth grade, we were doing long division. And not of functions.

 

[flyingpig]

No guys, I already know the correct method to get the answer explained in post #4. My question is why doesn't that other method work.

 

[eumyang]

But the other method DOES work. You're just not finished yet!

If $$(\frac{a}{b+a})^-1$$ = $$\frac{b+a}{a}$$ indeed, then couldn't I have done this

$$\begin{aligned}<br /> V^-1 &= P^-1(\frac{a}{b+a})^-1 \\<br /> \frac{P}{V} &= \frac{b+a}{a} \\<br /> \frac{P}{V} &= \frac{b}{a} + \frac{a}{a} \\<br /> \frac{P}{V} &= \frac{b}{a} + 1 \\<br /> \frac{P}{V} - 1 &= \frac{b}{a} \\<br /> a &= \frac{b}{\frac{P}{V} - 1}<br /> \end{aligned}$$

Clearly this is wrong

(Cleaned your LaTeX above.)

Now multiply the fraction by V/V:
$$\begin{aligned}<br /> a &= \frac{b}{\frac{P}{V} - 1} \cdot \frac{V}{V} \\<br /> &= \frac{Vb}{P - V}<br /> \end{aligned}$$

 

[zketrouble]

Aside from that, since when is this fifth grade? When I was in fifth grade, we were doing long division. And not of functions.

My thoughts exactly. This was the kind of math that I was doing in my first college math class Algebra II, MAT120, and it counted for college credit. Granted I was only 16 when I started college and I otherwise would've been in 10th grade to take it, but still this is WAAAAYYY ahead of 5th grade math.

 

[flyingpig]

But the other method DOES work. You're just not finished yet!


(Cleaned your LaTeX above.)

Now multiply the fraction by V/V:
$$\begin{aligned}<br /> a &= \frac{b}{\frac{P}{V} - 1} \cdot \frac{V}{V} \\<br /> &= \frac{Vb}{P - V}<br /> \end{aligned}$$

You know when I wrote this on paper, it looked completely different...

Are we treating (b-a)^-1 as x^-1? Because it cannot be b^-1 - a^-1

 

[Mark44]

(b - a)^-1 = 1/(b - a), but
1/(b - a) $\neq$ 1/b - 1/a