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- Sep 9th 2009, 12:05 AMrealisticsolve for y help
- Sep 9th 2009, 12:26 AMdhiab
- Sep 9th 2009, 12:29 AMpacman
-a/-y = x/a; solve for y?

cross-multiply,

-(a)^2 = -xy, cancel the negative sign

y = a^2/x

or

x = a^2/y

is this what you want? - Sep 9th 2009, 12:36 AMrealistic
- Sep 9th 2009, 12:38 AMrealistic
- Sep 9th 2009, 12:49 AMdhiab
HELLO :

$\displaystyle \begin{array}{l}

\frac{{ - y}}{{ - a}} = \frac{a}{x} \\

\left( { - y} \right) \times \left( x \right) = \left( { - a} \right) \times \left( a \right) \\

\left( { - y} \right) \times \left( x \right) = - a^2 \\

y \times x = a^2 \\

\end{array}$

divide by x :

$\displaystyle \frac{{y \times x}}{x} = \frac{{a^2 }}{x}

$

conclusion $\displaystyle y = \frac{{a^2 }}{x}

$ - Sep 9th 2009, 12:57 AMrealistic
- Sep 9th 2009, 12:57 AMfuror celtica
a negative times a negative equals a positive, man did you ever see stand and deliver or wot?

- Sep 9th 2009, 01:01 AMrealistic
- Sep 9th 2009, 01:08 AMfuror celtica
think of it this way

(-y) x (X) = -a^2

could be rewritten

-1 x y x X = -1 x a x a

-1 is a factor of both sides

so cut it out

(Evilgrin) - Sep 9th 2009, 01:12 AMrealistic
- Sep 9th 2009, 02:43 AMfuror celtica
sorry kid i cant understand your writing (is that paint or what?)

- Sep 9th 2009, 02:47 AMrealistic
- Sep 9th 2009, 02:58 AMdhiab
Hello : question 2

$\displaystyle

\begin{array}{l}

x = k - \frac{{\sqrt {pq} }}{y} \\

\frac{{\sqrt {pq} }}{y} = k - x \\

y = \frac{{\sqrt {pq} }}{{k - x}} \\

\end{array}

$(Rofl) - Sep 9th 2009, 03:20 AMrealistic