*Simultaneous equation 4 variables

[karush]

Solve the following system for x and y in terms of a and b where $ab\ne - 1$

$\left\{{\frac{x+y-1}{x-y+1}=a
\atop\frac{y-x+1}{x-y+1}=ab}\right.$

Answer is
$x=\frac{a+1}{ab+1}\\y=\frac{a\left(b+1\right)}{\left(ab+1\right)}$

I tried for a hour and?
Noticed the denominators were = but?

 

[topsquark]

Solve the following system for x and y in terms of a and b where $ab\ne - 1$

$\left\{{\frac{x+y-1}{x-y+1}=a
\atop\frac{y-x+1}{x-y+1}=ab}\right.$

Answer is
$x=\frac{a+1}{ab+1}\\y=\frac{a\left(b+1\right)}{\left(ab+1\right)}$

I tried for a hour and?
Noticed the denominators were = but?

From the first equation
x + y - 1 = a(x - y + 1)
or
(1 - a)x + (a + 1)y = a + 1

From the second equation:
y - x + 1 = ab(x - y + 1)
or
-(ab + 1)x + (ab + 1)y = ab - 1

If you don't want to match coefficients or use substitution you could always do it using Cramer's rule:
\(\displaystyle \left ( \begin{matrix} 1 - a & a + 1 \\ -(ab + 1) & ab + 1 \end{matrix} \right ) ~ \left ( \begin{matrix} x \\ y \end{matrix} \right ) = \left ( \begin{matrix} a + 1 \\ ab - 1 \end{matrix} \right )\)

So
\(\displaystyle x = \frac{ \left | \begin{matrix} a + 1 & a + 1 \\ ab - 1 & ab + 1 \end{matrix} \right | }{ \left | \begin{matrix} 1 - a & a + 1 \\ -(ab + 1) & ab + 1 \end{matrix} \right | }\)

and
\(\displaystyle y = \frac{ \left | \begin{matrix} 1 - a & a + 1 \\ -(ab + 1) & ab - 1 \end{matrix} \right | }{ \left | \begin{matrix} 1 - a & a + 1 \\ -(ab + 1) & ab + 1 \end{matrix} \right | }\)

It might look bad, but it's more "mechanized" than the other two ways, and easier to check.

-Dan