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I have a system of equations problem with $5$ variables but only $4$ equations:

Suppose that $x,y,z,u$ and $v$ are real numbers that satisfy the following system: $$ \begin{align} -4x + 6y + 14z + 4v &= -8\\ -3x + 9y + 15z - u + 3v &= 6\\ -3x + 6z + u + 3v &= -18\\ -x + 2y + 4z + v &= -1\\ \end{align} $$ Which of the following statements are true?

i. $y + z = 2$

ii. $v = 0$

iii. $x - 2z - v = 5$

iv. $y + z - u = 5$

Ans: The correct statements are: __

A. i. and iii.

B. i., ii., and iii.

C. i., iii., and iv.

D. i. and ii.

E. All

How should I approach the problem considering there are more variables than equations?

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closed as off-topic by Did, Andrew D. Hwang, user63181, user91500, azimut Apr 13 '14 at 14:41

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    $\begingroup$ Irrespective of numbers of variables and equations, a way to approach such problems is to set up the augmented matrix representing the system of equations and then use elementary row operations to bring it to reduced row-echelon form; from there it will probably be easy to see what the answer is. $\endgroup$ – Gerry Myerson Apr 13 '14 at 11:17
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Build the matrix of the system and apply Gaussian elimination: \begin{align} \left[\begin{array}{ccccc|c} −4 & 6 & 14 & 0 & 4 & −8\\ −3 & 9 & 15 & −1 & 3 & 6\\ −3 & 0 & 6 & 1 & 3 & −18\\ −1 & 2 & 4 & 0 & 1 &−1 \end{array}\right] &\to \left[\begin{array}{ccccc|c} −1 & 2 & 4 & 0 & 1 &−1 \\ −4 & 6 & 14 & 0 & 4 & −8\\ −3 & 9 & 15 & −1 & 3 & 6\\ −3 & 0 & 6 & 1 & 3 & −18 \end{array}\right]\text{ (swap 1 and 4)}\\ &\to \left[\begin{array}{ccccc|c} 1 & -2 & -4 & 0 & -1 & 1 \\ 0 & -2 & -2 & 0 & 0 & −4\\ 0 & 3 & 3 & −1 & 0 & 9\\ 0 & -6 & -6 & 1 & 0 & −15 \end{array}\right]\text{ (first column)}\\ &\to \left[\begin{array}{ccccc|c} 1 & -2 & -4 & 0 & -1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 2\\ 0 & 0 & 0 & −1 & 0 & 3\\ 0 & 0 & 0 & 1 & 0 & −3 \end{array}\right]\text{ (second column)}\\ &\to \left[\begin{array}{ccccc|c} 1 & -2 & -4 & 0 & -1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 2\\ 0 & 0 & 0 & 1 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]\text{ (fourth column)}\\ &\to \left[\begin{array}{ccccc|c} 1 & 0 & -2 & 0 & -1 & 5 \\ 0 & 1 & 1 & 0 & 0 & 2\\ 0 & 0 & 0 & 1 & 0 & -3\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]\text{ (second column)}\\ \end{align}

Your system has been reduced to the form \begin{cases} x=2z+v+5\\ y=-z+2\\ u=-3 \end{cases} Can you answer the questions now?

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  • $\begingroup$ Yes thanks! I was worried it had something to do with infinite equations. This is elegantly done, thanks. $\endgroup$ – castielle Apr 13 '14 at 14:18

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