Algebra Help: Solve Trig for x, y with Point of h Base

[raining Pete]

Hello all, was wondering if you fine chaps / ladies could help with a bit of algebra. Annoyed it's had to come to this but alas my algebra levels keep leading me astray.

$x + y = c$
$x^2 + h^2 = a^2$
$y^2 + h^2 = b^2$

I've been reading Measurements - Paul Lockhart (wish i picked this up 10 years ago) and the swine asks to first find x and y 'see if you can rearrange the equations to get':

$x = c/2 + (a^2 - b^2)/2c$
$y = c/2 - (a^2 - b^2)/2c$

After this it is to find h but i would like know the algebra behind finding x and y. I've swapped, rearranged and been all over the place, never the above though. For example,

$x^2 - y^2 = a^2 - b^2$ ==>
$-x(x-1) + y^2 + y = c - a^2 - b^2$ ==> etc,

trying quadratics, generally leading to nowhere or something else entirely.

Thanks for any help if given.

 

[Bystander]

Hint: x² - y² = (x+y)(x-y)

 

[raining Pete]

Hint: x² - y² = (x+y)(x-y)

Oh man, he was going on about Babylonian differences of squares three pages back.

Cheers Bystander.

 

[mathman]

Hello all, was wondering if you fine chaps / ladies could help with a bit of algebra. Annoyed it's had to come to this but alas my algebra levels keep leading me astray.

$x + y = c$
$x^2 + h^2 = a^2$
$y^2 + h^2 = b^2$

I've been reading Measurements - Paul Lockhart (wish i picked this up 10 years ago) and the swine asks to first find x and y 'see if you can rearrange the equations to get':

$x = c/2 + (a^2 - b^2)/2c$
$y = c/2 - (a^2 - b^2)/2c$

After this it is to find h but i would like know the algebra behind finding x and y. I've swapped, rearranged and been all over the place, never the above though. For example,

$x^2 - y^2 = a^2 - b^2$ ==>
$-x(x-1) + y^2 + y = c - a^2 - b^2$ ==> etc,

trying quadratics, generally leading to nowhere or something else entirely.

Thanks for any help if given.

$x^2-y^2=a^2-b^2$
$x^2-y^2=(x+y)(x-y)=c(x-y)$
therefore $x-y=\frac{a^2-b^2}{c}$

You can easily get x and y.