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.

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- Oct 22nd 2013, 06:38 AMLeatherneckTransposition 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. - Oct 22nd 2013, 07:04 AMSlipEternalRe: 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. - Oct 22nd 2013, 08:03 AMHallsofIvyRe: 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. - Oct 23rd 2013, 12:20 AMLeatherneckRe: Transposition of Formula
Not sure,I need practice.Thanks anyway

- Oct 23rd 2013, 12:21 AMLeatherneckRe: Transposition of Formula
Thanks