1. ## Transposition of Formula

Hi,

Could some one show me how to do this,its all new to me so as simple as possible please,reasoning behind it etc.
Thanks guys.

2. ## Re: Transposition of Formula

I am assuming the ones that are not circled are not due, so I can help with one of those. Let's look at problem 35.

$\displaystyle y^2 = 4a\left(x + \dfrac{c^2}{4a}\right)$

Let's read what is happening. On the left-hand side (LHS) of the equation, you have $\displaystyle y^2$. On the right-hand side (RHS), you have $\displaystyle 4a$ times an expression containing $\displaystyle x$. We want to isolate the expression containing $\displaystyle x$. We want that coefficient to be 1. So, we divide both sides by 4a:

$\displaystyle \dfrac{y^2}{4a} = x + \dfrac{c^2}{4a}$

We still want to get $\displaystyle x$ alone. Currently, on the LHS of the equation, we don't have any terms with $\displaystyle x$, but on the RHS, we have $\displaystyle x$ plus some term that does not contain any $\displaystyle x$. So, we subtract $\displaystyle \dfrac{c^2}{4a}$ from both sides of the equation:

$\displaystyle \dfrac{y^2}{4a} - \dfrac{c^2}{4a} = x$

Now we are done. We have $\displaystyle x$ on one side of the equation and everything that does not have an $\displaystyle x$ on the other.

3. ## Re: Transposition of Formula

To solve $\displaystyle \dfrac{x^2}{P^2}+ \dfrac{y^2}{q^2}= 1$ for P, we note that P is only in the second fraction so we can simplify by subtracting the other fraction from both sides: $\displaystyle \dfrac{x^2}{P^2}= 1- \dfrac{y^2}{q^2}$. At this point there are a couple of different ways to continue.

1) Probably simplest for this problem. Multiply both sides by $\displaystyle P^2$: $\displaystyle x^2= (1- \dfrac{y^2}{q^2})P^2$. That gets P out of the denominator. We still have $\displaystyle P^2$ multiplied by $\displaystyle 1- \dfrac{y^2}{q^2}$. To undo that, we must divide both sides by $\displaystyle 1- \dfrac{y^2}{q^2}= \dfrac{q^2- y^2}{q^2}$. That gives [tex]\dfrac{q^2x^2}{q^2- y^2}= P^2[/itex]. The last step, of course, would be to take the square root of both sides.

1) Quicker but harder to see. Start by writing the $\displaystyle 1- \dfrac{y^2}{q^2}$ as $\displaystyle \dfrac{q^2- y^2}{q^2}$ so your equation is $\displaystyle \dfrac{x^2}{P^2}= \dfrac{q^2- y^2}{q^2}$. Now, "cross multiply"- that is, multiply both sides by $\displaystyle P^2q^2$ to get rid of the two denominators: $\displaystyle x^2q^2= P^2(q^2- y^2)$. Get $\displaystyle P^2$ alone on the right by dividing both sides by [tex]q^2- y^2[tex]. That leaves $\displaystyle \dfrac{x^2q^2}{q^2- y^2}= P^2$ and now you can take the square root of both sides.

The general idea is this: You want to "solve" and equation for a given variable because it is NOT "by itself" on one side of the equation. Things have been done to it- it has been multiplied by something or something has been added or subtracted or it has been squared or cubed, etc. To get it by itself, you have to undo those things- do the opposite. And, of course, always do the same thing to both sides of the equation.

4. ## Re: Transposition of Formula

Not sure,I need practice.Thanks anyway

Thanks