# Thread: Solving for variable in expression

1. ## Solving for variable in expression

Hi,

I have the following expression:

$\displaystyle \frac{V^2(R_t +R_L)^2 -2V^2R_L(R_t + R_L)} {(R_t+R_L)^4}=0$

I need to solve for $\displaystyle R_L$

If you multiply the whole expression by: $\displaystyle (R_L + R_t)^4$
That part just disappears?

Maybe it's easier to just cancel out the (R_t +R_L) parts?
So you have $\displaystyle \frac{(V^2-2V^2R_L)}{(R_t + R_L)}=0$
but then what?

//Jones

2. Originally Posted by Jones
$\displaystyle \frac{V^2(R_t +R_L)^2 -2V^2R_L(R_t + R_L)} {(R_t+R_L)^4}=0$

I need to solve for $\displaystyle R_L$

If you multiply the whole expression by: $\displaystyle (R_L + R_t)^4$
That part just disappears?
By multiplying on both sides, the denominator "cancels" on the left-hand side; anything times zero is just zero, so that's why nothing changes on the right-hand side.

You have:

$\displaystyle V^2(R_t\, +\,R_L)^2 \,-\,2V^2R_L(R_t \,+\, R_L)\, =\, 0$

You can also divide through by V^2:

$\displaystyle (R_t\, +\,R_L)^2 \,-\,2R_L(R_t \,+\, R_L)\, =\, 0$

Then square things out:

$\displaystyle R_t^2\,+\,2R_t R_L\,+\,R_L^2\,-\,2 R_L R_t \,-\,2 R_L^2\,=\,0$

$\displaystyle 0\, =\, R_L^2\, -\, R_t^2$

$\displaystyle 0\, =\, (R_L\, -\, R_t)(R_L\, +\, R_t)$

Then complete the solution for R_L in terms of R_t.

3. More fundamentally, the only way a fraction can be equal to 0 is if the numerator is 0- it doesn't matter what the denominator is: $\displaystyle \frac{p}{q}= 0$ if and only if p= 0. As far as the $\displaystyle (R_t+ R_L)$ in the numerator is concerned, the fraction is only defined if it is not 0. In that case it doesn't matter whether you cancel before multiplying both sides by the denominator or not.