Question 865898
<pre>
3x-4y+2z+8=0, x+5y-3z+2=0, 5x+3y-z+6=0

Get the constants on the right:

3x-4y+2z=-8, x+5y-3z=-2, 5x+3y-z=-6

I'll color the constants red:

{{{system(3x-4y+2z=red(-8),x+5y-3z=red(-2),5x+3y-z=red(-6))}}}

{{{D=abs(matrix(3,3,
3,-4, 2,
1, 5, -3,
5, 3, -1))}}} 

x is the FIRST unknown and so {{{D[x]}}} is just like {{{D}}}
except that the FIRST column is replaced by the three red numbers: 

{{{D[x]=abs(matrix(3,3,
red(-8),-4, 2,
red(-2), 5, -3,
red(-6), 3, -1))}}}

y is the SECOND unknown and so {{{D[y]}}} is just like {{{D}}}
except that the SECOND column is replaced by the three red numbers: 


{{{D[y]=abs(matrix(3,3,
3,red(-8), 2,
1, red(-2), -3,
5, red(-6), -1))}}}

z is the THIRD unknown and so {{{D[z]}}} is just like {{{D}}}
except that the THIRD column is replaced by the three red numbers,

{{{D[z]=abs(matrix(3,3,
3,-4, red(-8),
1, 5, red(-2),
5, 3, red(-6)))}}}

Do you know how to evaluate a 3x3 determinant?
If not, post again asking how to.  I will assume
you already know how.

{{{D=24}}}
{{{D[x]=-48}}}
{{{D[y]=72}}}
{{{D[z]=120}}}

Then the solutions are:

{{{x=D[x]/D=(-48)/24 = -2}}}
{{{y=D[y]/D=(72)/24 = 3}}}
{{{z=D[z]/D=(120)/24 = 5}}}

Edwin</pre>