# help solving 3 algebraic statements simultaneously

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.