How to solve for 3 variables with 2 equations? What is a systematic way of determining if a system of equations has a unique solution? For example, this system with $2$ equations and $3$ unknowns: $$2a+b+c=0\\
6a+2b+c=0.$$
 A: With only those two equations, there is not a unique solution for $a$, $b$, and $c$.
Subtracting the first equation from the second yields $4a+b=0$ or $b=-4a$.
Substituting $-4a$ for $b$ in either equation yields $-2a+c=0$ or $c=2a$.
Can you verify that, if $a$ is any real number,
then with $b=-4a$ and $c=2a$ both equations are satisfied?
A: *

*Let a system of equations have $q$ linearly-independent
equations and $n$ unknowns.

*

*If it has infinitely many solutions, i.e., is undetermined,    then $q<n.$

*If it has a unique solution,then $q=n.$

*If $q>n,$ then the system has no solution, i.e., is inconsistent, i.e., is overdetermined.

Note that in each of the above cases, the converse is false.


*Your given system has only two equations; it is easy to verify that
it is indeed linearly independent. Since it has $q=2\ne3=n,$ it has no
unique solution.
A: Like any system of equations, you can subtract the equation on the bottom from the one on the top to remove a variable from both equations. In this case, we can subtract c. You will be left with just -4a-b=0 or, more cleanly, 4a+b=0. It is possible to formulate this in terms of a linear equation b=-4a. So no, there is no one answer. There is one b for every a.
A: You can imagine your equations as an equations of planes in 3d (where a is x, b is y and c is z).
If you'd had 3 linear equations - 3 planes and not parallel to each-other, you would get an intersection - one point => single soultion.
When you have two planes, they give an intersection as a line.
Basically, this line in 3d space is your solution.
Look here for visualization: https://www.math3d.org/xIOmOlVN
