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This is like a fifth grade question

  • Thread starter flyingpig
  • Start date
  • #1
2,571
1

Homework Statement





Let's say I have [tex]V = P\frac{a}{b+a}[/tex]

Solve for a

So I did the math and I got [tex]\frac{Vb}{V+P}[/tex]

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

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

Answers and Replies

  • #2
Pengwuino
Gold Member
4,989
15
For one, your 2nd line is incorrect, check your math.

Second, why should they be equal?
 
  • #3
33,493
5,183

Homework Statement





Let's say I have [tex]V = P\frac{a}{b+a}[/tex]

Solve for a

So I did the math and I got [tex]\frac{Vb}{V+P}[/tex]
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 [tex]V = P\frac{a}{b+a}[/tex] to [tex]V^-1 = P^-1(\frac{a}{b+a})^-1[/tex]

Now my question is, why isn't [tex](\frac{a}{b+a})^-1[/tex] = [tex]\frac{b+a}{a}[/tex]
It is, but that doesn't do you much good if you're trying to solve for a.
 
  • #4
2,571
1
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?
 
  • #5
Mentallic
Homework Helper
3,798
94
Yes, [tex]\left(\frac{a+b}{a}\right)^{-1}=\frac{a}{a+b}[/tex] as you stated.
 
  • #6
47
0

Homework Statement





Let's say I have [tex]V = P\frac{a}{b+a}[/tex]

Solve for a

So I did the math and I got [tex]\frac{Vb}{V+P}[/tex]

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

Now my question is, why isn't [tex](\frac{a}{b+a})^-1[/tex] = [tex]\frac{b+a}{a}[/tex]
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?
 
  • #7
2,571
1
If [tex](\frac{a}{b+a})^-1[/tex] = [tex]\frac{b+a}{a}[/tex] indeed, then couldn't I have done this

[tex]V^-1 = P^-1(\frac{a}{b+a})^-1[/tex]

[tex]\frac{P}{V} = \frac{b+a}{a}[/tex]


[tex]\frac{P}{V} = \frac{b}{a} + \frac{a}{a}[/tex]



[tex]\frac{P}{V} = \frac{b}{a} + 1[/tex]

[tex]\frac{P}{V} = \frac{b}{a} + 1[/tex]

[tex]\frac{P}{V} - 1 = \frac{b}{a} [/tex]

[tex] a = \frac{b}{]\frac{P}{V} - 1} [/tex]

Clearly this is wrong
 
  • #8
326
1
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)
 
  • #9
Char. Limit
Gold Member
1,204
14
Aside from that, since when is this fifth grade? When I was in fifth grade, we were doing long division. And not of functions.
 
  • #10
2,571
1
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.
 
  • #11
eumyang
Homework Helper
1,347
10
But the other method DOES work. You're just not finished yet!

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

[tex]\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}[/tex]

Clearly this is wrong
(Cleaned your LaTeX above.)

Now multiply the fraction by V/V:
[tex]\begin{aligned}
a &= \frac{b}{\frac{P}{V} - 1} \cdot \frac{V}{V} \\
&= \frac{Vb}{P - V}
\end{aligned}[/tex]
 
  • #12
47
0
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.
 
  • #13
2,571
1
But the other method DOES work. You're just not finished yet!


(Cleaned your LaTeX above.)

Now multiply the fraction by V/V:
[tex]\begin{aligned}
a &= \frac{b}{\frac{P}{V} - 1} \cdot \frac{V}{V} \\
&= \frac{Vb}{P - V}
\end{aligned}[/tex]
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
 
  • #14
33,493
5,183
(b - a)-1 = 1/(b - a), but
1/(b - a) [itex]\neq[/itex] 1/b - 1/a
 

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