The expression

is a linear equation, with coefficients $a_1=2$, $a_2=3$, $a_3=4$ and independent term $b=24$. When few unknowns are used, it is customary to call them $x$, $y$, $z$, $t$... just to avoid many subindexes

For instance $(x=1,y=2,z=4)$ is a solution of this linear equation, because replacing the unknowns with the chosen values satisfies the equality

$(x=10,y=-8,z=7)$ is a solution too

but $(x=1,y=1,z=1)$ is not a solution

that is, some choices may be solutions but others may not. We would like to find *all* solutions

Now we may add a second equation

thus having a system of linear equations. For this system, our first choice $(x=1,y=2,z=4)$ is still a solution:

but the second choice $(x=10,y=-8,z=7)$ is no longer a solution

This is kind of obvious: the more equations we have to satisfy, the less solutions we find. As before, we would like to find *all* solutions - how should we proceed?