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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

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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.

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$s$ and $t$ are representing constants, but $x_2$ and $x_3$ are variables. Meaning is "if we fix this terms ...". Daniel

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There is no real necessity to introduce new variables.

They may be useful though, as they can help pointing out that the solutions are in some particular form and depend only on one or some arbitrary variable(s).

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 $x=t$.

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