1. ## difficult algebra problem

If (a^2) + bc = 1 (a, b, c are given real numbers) , show that

(cx - y(1+a))/(by - x(1-a)) = -(1+a)/b = -c/(1-a).
where (x and y are any real numbers)

I thought factoring: bc = 1 - (a^2) = (a+1)(a-1).

From this it is easily shown that -(1+a)/b = -c/(1-a).

But how to show that any of these equals: (cx - y(1+a))/(by - x(1-a)).

It all seems quite intriguing to me. Does anyone see any light here?

2. Hello,

Factorize above by 1+a and below by b.

$= \frac{1+a}{b} ( \frac{\frac{c}{1+a} x - y}{y-\frac{1-a}{b} x} )$

As you have noticed, bc=(1+a)(1-a)

Hence $\frac{c}{1+a}=\frac{1-a}{b}$

-> $= \frac{1+a}{b} ( \frac{\frac{1-a}{b} x - y}{y-\frac{1-a}{b} x} ) = - \frac{1+a}{b}$

Is it clear ? :-)

3. Thanks moo,
All clear.