# Finding the solution set for equations

Given an equation: $3x-4y=-1$, if I set $x=t$, and solve for $y$, I end up with $y = {1 \over 4} + {3 \over 4}t$ and the solution set is $[{1 \over 4} + {3 \over 4}t, t]$

Similarly if my equation is $x_{1} - x_{2} - 2x_{3} = 3$ and I set $x_{2}=s, x_{3} = t$, and solve for $x_{1}$ then the solution set is $[3+s-2t, s, t]$

In each example I understand or at least I think it makes sense why the first term is in the solution sets is part of the answer (because that's what $y$ and $x_{1}$ equal after some algebra). What I don't understand is why do you need additional terms for each variable (i.e the $t$ in the first example and the $s$, $t$ in the second example.

if anyone is wondering this is from the book: Linear Algebra A modern Introduction Third edition by David Poole

The solution sets consist of tuples: in the first case, $(x,y)$ pairs; in the second, $(x_1,x_2,x_3)$ triples. $\frac14+\frac34t$ is a number, not an ordered pair, so you still need an $x$-value to go with this $y$-value. In the same way, $3+s-2t$ is a number, not an ordered triple, so you still need $x_2$ and $x_3$ components for the triples that are in the solution set.
$s$ and $t$ are representing constants, but $x_2$ and $x_3$ are variables. Meaning is "if we fix this terms ...". Daniel
In some occasions they can also simplify the work: (trivial example) in $x-\sqrt{y}=0$ putting $y=t^2$ (with the obvius condition the $y$ be non negative) you get $x=t$.