I was wondering if there is a unique solution for a, b and c.
I can't seem to come up with one with both equations equaling to 0.
It would be greatly appreciated for some suggestions.
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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?
Let a system of equations have $q$ linearly-independent equations and $n$ unknowns.
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.
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
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.