Having huge troubles with this whole section

$\displaystyle

y^2-9x^2=19$

$\displaystyle y=-5x$

Also, does anyone know any good math books for post-Algebra II? Math is cool when I get it, which is why I want to get it.

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- Mar 15th 2009, 01:01 PM #1

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- Mar 15th 2009, 02:02 PM #2

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the hard part is done for you. you have two equations, and one of them tells you one variable in terms of the other, without any additional manipulation.

you take y=-5x and plug that into the first equation. you have two choices with this:

(-5x)^2-9x^2=19

or, if you solve for x in the second equation (and get x=-(1/5)y)

y^2-9(-(1/5)y)^2=19

the first way is easier so lets stick to that.

(-5x)^2-9x^2=19

25x^2-9x^2=19

16x^2=19

x^2=19/16

x=sqrt(19/16).

now take this value for x and plug it into the second equation of our original pair.

y=-5sqrt(19/16)

hope this helps.

- Mar 15th 2009, 02:23 PM #3

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- Mar 15th 2009, 10:48 PM #4

Hi B-Ren & Robeuler,

We will substitute y from

y= -5x

into the second equation

The second equation thus becomes

$\displaystyle (-5x)^2 - 9x^2 = 19$

$\displaystyle (-5)^2 \times (x)^2 - 9 x^2 = 19 $

$\displaystyle 25x^2 - 9x^2 = 19$

$\displaystyle 16x^2 = 19$

$\displaystyle x^2 = \frac{16}{19} $

$\displaystyle x = \pm \sqrt{\frac{16}{19}} $ ---->**An important Discovery**

Putting this in first equation we get

$\displaystyle y = -5x$

$\displaystyle \implies y = (-5)\times(\pm\sqrt{\frac{16}{19}}) $

- Mar 16th 2009, 09:44 AM #5

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