1. ## Suptitution

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.

2. Originally Posted by B-Ren
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.
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.

3. Originally Posted by robeuler
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.
How is it that the 25x retains its ^2?

4. Originally Posted by B-Ren
How is it that the 25x retains its ^2?

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

$\displaystyle y = -5x$

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

5. Originally Posted by B-Ren
How is it that the 25x retains its ^2?
It doesn't: x "retains its ^2". (5x)^2 means that both 5 and x are squared: $\displaystyle 5^2= 25$ but there is no simpler way to write $\displaystyle x^2$.