MHB Algebraic expression simplification 2(2u+v)/u = 4(v^2-4u^2)/3u^2

[Nightspider]

Hi All

I am trying to simplify an algebraic expression, I know what the expression should simplify to but I am struggling to simplify it.

The expression is: 2(2u+v)/u = 4(v^2-4u^2)/3u^2

and it is supposed to simplify to: 3u = 2(v-2u)

I have approached this problem from various angles by rearranging the expression and factorization but I can't seem to simplify it. Any pointers would be greatly appreciated. This is my first time posting on a forum so if I have not followed any of the rules of the forum please advise me. In the interest of transparency this is part of an assignment question from a module in a part time Physics degree I am doing as a mature student. For the purposes of this assignment I believe the important thing is to understand the Physics hence the fact that I am asking for some help with the math's. That being said I spent a significant amount of time trying to understand the math's before posting here.

Many Thanks

 

[Joppy]

Hi All

I am trying to simplify an algebraic expression, I know what the expression should simplify to but I am struggling to simplify it.

The expression is: 2(2u+v)/u = 4(v^2-4u^2)/3u^2

and it is supposed to simplify to: 3u = 2(v-2u)

I have approached this problem from various angles by rearranging the expression and factorization but I can't seem to simplify it. Any pointers would be greatly appreciated. This is my first time posting on a forum so if I have not followed any of the rules of the forum please advise me. In the interest of transparency this is part of an assignment question from a module in a part time Physics degree I am doing as a mature student. For the purposes of this assignment I believe the important thing is to understand the Physics hence the fact that I am asking for some help with the math's. That being said I spent a significant amount of time trying to understand the math's before posting here.

Many Thanks

So we have, $ \dfrac{2(v + 2u)}{u} = \dfrac{4(v^2-4u^2)}{3u^2}$. If we multiply both sides by $3u^2$ we obtain,

$ 6u(v + 2u)= 4(v^2-4u^2)$.

Now the right hand side can be expressed as: $4(v^2 - (2u)^2) = 4(v - 2u)(v + 2u)$. This is called the difference of two squares, and when first starting out, it isn't necessarily an obvious step.

Continuing,

$ 6u(v + 2u)= 4(v - 2u)(v + 2u)$,

now if we divide by $2(v + 2u)$ we arrive at,

$ 3u = \dfrac{\cancelto{2}{4}(v - 2u) \cancel{(v + 2u)}}{\cancel{2} \cancel{(v + 2u)}}$ which is desired expression,

$3u = 2(v - 2u)$.

 

[HallsofIvy]

Notice that this equivalence is true only if [math]v+ 2u\ne 0[/math].

 

[Nightspider]

Brilliant many thanks. Seems easy once someone who knows how shows you but I spent quite some time looking at it. Your swift reply is very much appreciated.

Thank You

 

[Greg]

$7u=2v$