# Use Gaussian elimination to find the complete solution to the system of given equations, or show that none exists{(5x,+,8y,-,6y,=,14),(3x,+,4y,-,2z,=,8),(x,+,2y,-,2z,=,3):}

Use Gaussian elimination to find the complete solution to the system of given equations, or show that none exists
$\left\{\begin{array}{l}5x+8y-6y=14\\ 3x+4y-2z=8\\ x+2y-2z=3\end{array}$

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consider the following system of linear equations:
$5x+8y-6z=14$
$3x+4y-2z=8$
$x+2y-2z=3$
convert into augmented matrix
$\left[\begin{array}{ccccc}5& 8& -6& |& 14\\ 3& 4& -2& |& 8\\ 1& 2& -2& |& 3\end{array}\right]$
Transform the above matrix into reduced row echelon form
${R}_{2}⇒\frac{5}{3}{R}_{2}-{R}_{1}$
${R}_{3}⇒5{R}_{3}-{R}_{1}$
$\left[\begin{array}{ccccc}5& 8& -6& |& 14\\ 0& -\frac{4}{3}& -\frac{8}{3}& |& -\frac{2}{3}\\ 0& 2& -4& |& 1\end{array}\right]$
${R}_{3}⇒\left(\frac{2}{3}\right){R}_{3}+{R}_{2}$
$\left[\begin{array}{ccccc}5& 8& -6& |& 14\\ 0& -\frac{4}{3}& -\frac{8}{3}& |& -\frac{2}{3}\\ 0& 0& 0& |& 0\end{array}\right]$
Hence, solution of a system of linear equations does not exist