solve for y help

**dhiab** · September 9th 2009, 12:26 AM

Originally Posted by realistic

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

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

**pacman** · September 9th 2009, 12:29 AM

-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?

**realistic** · September 9th 2009, 12:36 AM

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

**realistic** · September 9th 2009, 12:38 AM

Originally Posted by dhiab

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

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

ya thanks... can you help me with the second one plz

**dhiab** · September 9th 2009, 12:49 AM

Originally Posted by realistic

ya thanks... can you help me with the second one plz

HELLO :
$\begin{array}{l} \frac{{ - y}}{{ - a}} = \frac{a}{x} \\ \left( { - y} \right) \times \left( x \right) = \left( { - a} \right) \times \left( a \right) \\ \left( { - y} \right) \times \left( x \right) = - a^2 \\ y \times x = a^2 \\ \end{array}$
divide by x :

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

**realistic** · September 9th 2009, 12:57 AM

Originally Posted by dhiab

HELLO :
$\begin{array}{l} \frac{{ - y}}{{ - a}} = \frac{a}{x} \\ \left( { - y} \right) \times \left( x \right) = \left( { - a} \right) \times \left( a \right) \\ \left( { - y} \right) \times \left( x \right) = - a^2 \\ y \times x = a^2 \\ \end{array}$
divide by x :

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

could you pleae help me with the second question plz

**furor celtica** · September 9th 2009, 12:57 AM

a negative times a negative equals a positive, man did you ever see stand and deliver or wot?

**realistic** · September 9th 2009, 01:01 AM

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]

$ \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?

**furor celtica** · September 9th 2009, 01:08 AM

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

**realistic** · September 9th 2009, 01:12 AM

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

cool i see now clearly

also could you help out on the second question in the title plz

**furor celtica** · September 9th 2009, 02:43 AM

sorry kid i cant understand your writing (is that paint or what?)

**realistic** · September 9th 2009, 02:47 AM

Originally Posted by furor celtica

sorry kid i cant understand your writing (is that paint or what?)

yah paint

**dhiab** · September 9th 2009, 02:58 AM

Hello : question 2
$ \begin{array}{l} x = k - \frac{{\sqrt {pq} }}{y} \\ \frac{{\sqrt {pq} }}{y} = k - x \\ y = \frac{{\sqrt {pq} }}{{k - x}} \\ \end{array} $

**realistic** · September 9th 2009, 03:20 AM

Originally Posted by dhiab

Hello : question 2
$ \begin{array}{l} \frac{{\sqrt {pq} }}{y} = k - x \\ y = \frac{{\sqrt {pq} }}{{k - x}} \\ \end{array} $

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

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