# Thread: solve for y help

1. ## solve for y help

2. 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

3. -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?

4. 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

5. 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

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

7. 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

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

9. 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?

10. 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

11. 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

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

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

yah paint

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

15. Originally Posted by dhiab
Hello : question 2
$\displaystyle \begin{array}{l} \frac{{\sqrt {pq} }}{y} = k - x \\ y = \frac{{\sqrt {pq} }}{{k - x}} \\ \end{array}$
can you show the step you did in the middle of that? plz

Page 1 of 2 12 Last