# solve for y help

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• Sep 9th 2009, 12:05 AM
realistic
solve for y help
• Sep 9th 2009, 12:26 AM
dhiab
Quote:

Originally Posted by realistic

Hello: $\displaystyle \begin{array}{l} \frac{{ - y}}{{ - a}} = \frac{a}{x} \\ - y = \frac{{ - a^2 }}{x} \\ \end{array}$
Solve for y : $\displaystyle y = \frac{{a^2 }}{x}$

Solve for y :$\displaystyle x = \frac{{a^2 }}{y}$
THANK YOU
• Sep 9th 2009, 12:29 AM
pacman
-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 AM
realistic
Quote:

Originally Posted by pacman
-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?

can you explain how you cancelled the negative after cross multi? not sure how this happens
• Sep 9th 2009, 12:38 AM
realistic
Quote:

Originally Posted by dhiab
Hello: $\displaystyle \begin{array}{l} \frac{{ - y}}{{ - a}} = \frac{a}{x} \\ - y = \frac{{ - a^2 }}{x} \\ \end{array}$
Solve for y : $\displaystyle y = \frac{{a^2 }}{x}$

Solve for y :$\displaystyle x = \frac{{a^2 }}{y}$
THANK YOU

ya thanks... can you help me with the second one plz
• Sep 9th 2009, 12:49 AM
dhiab
Quote:

Originally Posted by realistic
ya thanks... can you help me with the second one plz

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 AM
realistic
Quote:

Originally Posted by dhiab
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 $\displaystyle y = \frac{{a^2 }}{x}$

could you pleae help me with the second question plz (Bow)
• Sep 9th 2009, 12:57 AM
furor celtica
a negative times a negative equals a positive, man did you ever see stand and deliver or wot?
• Sep 9th 2009, 01:01 AM
realistic
Quote:

Originally Posted by furor celtica
a negative times a negative equals a positive, man did you ever see stand and deliver or wot?

i know that i just cant see this step

[tex]

$\displaystyle \begin{array}{l} \left( { - y} \right) \times \left( x \right) = - a^2 \\ y \times x = a^2 \\ \end{array}$

are you just x it self by a negative?
• Sep 9th 2009, 01:08 AM
furor 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 AM
realistic
Quote:

Originally Posted by furor 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)

cool i see now clearly (Bow)

also could you help out on the second question in the title plz
• Sep 9th 2009, 02:43 AM
furor celtica
sorry kid i cant understand your writing (is that paint or what?)
• Sep 9th 2009, 02:47 AM
realistic
Quote:

Originally Posted by furor celtica
sorry kid i cant understand your writing (is that paint or what?)

yah paint

http://img200.imageshack.us/img200/1084/againo.jpg
• Sep 9th 2009, 02:58 AM
dhiab
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 AM
realistic
Quote:

Originally Posted by dhiab
Hello : question 2
$\displaystyle \begin{array}{l} \frac{{\sqrt {pq} }}{y} = k - x \\ y = \frac{{\sqrt {pq} }}{{k - x}} \\ \end{array}$(Rofl)

can you show the step you did in the middle of that? plz
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