i have three algebraic expressions, each using the others. in these equations a, b, c and t are known and plugged in later:
$x = a^{-1}(t + y + z)$
$y = b^{-1}(t + x + z)$
$z = c^{-1}(t + x + y)$
i have managed to successfully solve the equations for two statements using substitution:
$x = a^{-1}(t + y)$
$y = b^{-1}(t + x)$
which substitutes and simplifies as:
$x = a^{-1}t + a^{-1}b^{-1}t + a^{-1}b^{-1}x$
$x – a^{-1}b^{-1}x = a^{-1}t + a^{-1}b^{-1}t$
$x(1 – a^{-1}b^{-1}) = a^{-1}t(b^{-1} + 1)$
therefore:
$x = a^{-1}t(1+b^{-1}) / (1 – a^{-1}b^{-1})$
$y = b^{-1}t(1+a^{-1}) / (1 – a^{-1}b^{-1})$
but introducing a third term has me stumped, as i keep getting a nonsense result when substituting leaving me with $x = x$ when i simplify the equation. (this is obviously correct, but also useless.) I’m hoping that someone would be so kind as to explain how i go about substituting more than one term and solving x, y and z in terms of t, a, b and c. does the same principle apply as with two terms?
for completeness, here is my (incorrect) work:
$x = a^{-1}t + a^{-1}y + a^{-1}z$
$a^{-1}z = x – (a^{-1}t + a^{-1}y)$
$a^{-1}z = x – a^{-1}(t + y)$
$z = ax – (t + y)$ (multiply both sides by a)
therefore, to find y i substitute z:
$x = a^{-1}t + a^{-1}y + a^{-1}(ax – (t + y))$
$x = a^{-1}t + a^{-1}y + x – a^{-1}t – a^{-1}y$
$x = (a^{-1}t – a^{-1}t) + (a^{-1}y – a^{-1}y) + x$
$x = 0 + 0 + x$
$x = x$
please forgive me if my terminology isn’t correct. i don’t exactly have a maths background and english isn’t my first language.
