According to this note from the PrepSwift video on Systems of Equations, we need the same number of variables as the number of equations for the equations to be solvable. So in three equations, every equation must contain three variables, for example, x, y and z.
If that’s the case, I’m not sure why/how my answer (“No, unfortunately”) to the question below was wrong. There are three equations, but equation 2 has only two variables
6a + c = 40 is the same as 6a + 0b + c = 40.
Perhaps, you can use the following note:
A system of equations has a unique solution for n variables:
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You must have at least n equations
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At least n of those equations are linearly independent: As in, each equation provides unique information and are not just “scaled up” versions or combinations (sums or differences) of the other equations.
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The equations shouldn’t contradict each other.
For example, x + y = 2 and x + y = 5 are contradictions, so you can’t have both of them in the same system of equations.
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Thank you so much; the note makes it clearer. Just one last question, please, in what case would a system of equations with 3 equations and 3 variables be unsolvable?
According to the notes from above, a system is “unsolvable” (inconsistent) when at least one pair of equations contradicts each other.
For example, consider the following system:
\begin{cases}
2x + 10y + 6z = 8 \\
x + y + z = 6 \\
3x + 3y + 3z = 10
\end{cases}
As you can see, the third equation contradicts the second (WHY?). It is thus “unsolvable” because there is no (x, y, z) that can simultaneously satisfy all three equations.
The 3rd equation contradicts the second because, when broken down, they are both identical equations equal to different values (they will both become parallel lines), and the final answer will make no sense. Is this correct, please?
Thank you very much for your help so far.
Yes, something like that. The second equation is equivalent to 3x + 3y + 3z = 18, while the third equation suggests a different result: 3x + 3y + 3z = 10. It’s obvious that no solution (x,y,z) exists which satisfies both equations at the same time.
They’re planes not lines.