2
$\begingroup$
2a+b+c=0
6a+2b+c=0

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.

$\endgroup$
0
5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$
  1. 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.

  2. Your given system has only two equations; it is easy to verify that it is indeed linearly independent. Since it has $2=q<n=3,$ it has no unique solution.

$\endgroup$
2
  • $\begingroup$ what about the system $x=0, 2x=0, 3x=0$? $\endgroup$ Oct 22 at 2:55
  • 1
    $\begingroup$ @J.W.Tanner Thanks for the catch! Have corrected. $\endgroup$
    – ryang
    Oct 22 at 3:29
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Oct 22 at 4:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.