# Find values for $a$, $b$, $c$ that make this linear system solvable?

I came along with the following exercise that I developed poorly. May anybody give me some light? See:

How to find a solution involving $a$, $b$, $c$ to make the following system consistent? Find the solutions when possible.

$$x + y + z + t = a$$ $$5y + 2z + 4t = b$$ $$3x - 2y + z - t = c$$

Well. First I tried to reduce the system to the reduced row echelon form, with not very success. What I got is:

$$x - \frac{2}{5}t = \frac{5a - b}{5}$$ $$y + \frac{2}{5}t = \frac{-c + 3a}{5}$$ $$z + t = \frac{c - 3a + b}{5}$$

I thought it would help, but I don't know how to resume the exercise.

Any tips?

Thank you.

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You had better check you reduction, as what you got is not correct. – Gerry Myerson Oct 14 '12 at 11:54
You are right. I fixed the reduction. Thank you. – Silas Oct 14 '12 at 13:02

The very first step in row reduction gives the new equation $$-5y -2z -4t = c-3a$$ Since the left-hand side of this is minus the left-hand side of the equation for $b$, the equations imply $b=3a-c$.

On the other hand, that same first step of row reduction shows that the rank of the equation system is at least 2, so the column space is at least 2-dimensional. The condition $b=3a-c$ already restricts the possible $(a,b,c)$ to a 2-dimensional subspace, to there is no room for further restrictions.

So the possible $(a,b,c)$ are exactly those that satisfy $b=3a-c$ (or equivalently $3a-b-c=0$ or $b+c=3a$),

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Supposed that this is correct so far, I think, you can finish it by saying

Let $a,b,c$ be arbitrary, moreover let $t$ be arbitrary too. Then you get unique solution for $x,y,z$ as above.

In other words, for every $a,b,c$ there are infinitely many solutions (as $t$ is arbitrary)..

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