# This is like a fifth grade question

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

Related Precalculus Mathematics Homework Help News on Phys.org
Pengwuino
Gold Member

Second, why should they be equal?

Mark44
Mentor

## 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.

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

## 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?

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

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

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
Homework Helper
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} 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} - 1 &= \frac{b}{a} \\ a &= \frac{b}{\frac{P}{V} - 1} \end{aligned}

Clearly this is wrong

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

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.

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

Now multiply the fraction by V/V:
\begin{aligned} a &= \frac{b}{\frac{P}{V} - 1} \cdot \frac{V}{V} \\ &= \frac{Vb}{P - V} \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
Mentor
(b - a)-1 = 1/(b - a), but
1/(b - a) $\neq$ 1/b - 1/a