<PRE><B>Question 1090283</B>
&lt;br&gt;There are only two equations and three unknowns.  So if the system has a solution, it will be an infinite set of solutions defined by a parameter.&lt;br&gt;
Because of the way Gaussian reduction is performed, the solution will have x and y expressed in terms of parameter z.&lt;br&gt;&lt;br&gt;
The matrix for the given system of equations is
{{{matrix(2,4,2,-1,4,1,0,1,1,3)}}}&lt;br&gt;
I don&#39;t like introducing fractions into the matrix when doing Gaussian reduction; but in this case we have no choice.  So divide the first row by 2:&lt;br&gt;
{{{matrix(2,4,1,-1/2,2,1/2,0,1,1,3)}}}&lt;br&gt;
Next use the 1 in row 2 column 2 to get a 0 in row 1 column 2:&lt;br&gt;
R1 &lt;-- R1 + (1/2)R2: 1+0=1; -1/2+1/2=0; 2+1/2=5/2; 1/2+3/2=2.
{{{matrix(2,4,1,0,5/2,2,0,1,1,3)}}}&lt;br&gt;
This is as far as we can go with Gaussian reduction.  The final matrix gives us these equations:&lt;br&gt;
x+(5/2)z=2; y+z=3&lt;br&gt;
We rewrite these equations to give us parametric equations for x and y in terms of parameter z:&lt;br&gt;
x = 2-(5/2)z; y = 3-z&lt;br&gt;
And the parametric solution set is&lt;br&gt;
x = 2-(5/2)z;
y=3-z;
z=z


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