Question 1034917
<pre><font size = 3><b>
Instead of doing yours for you, I'll do one exactly
like it, so you can use it as a model and learn to
do this kind of problem by yourself:

x+y+z = 6
2x-y+z = 3
x+2y-3z = -4

First put in all the 1, -1, and 0 coefficients, if
necessary:

{{{system(1x+1y+1z=red(6),
2x-1y+1z=red(3),
1x+2y-3z=red(-4))}}}

Cramer's rule:
 
There are 4 columns,
 
1. The column of x-coefficients {{{matrix(3,1,1,2,1)}}}
 
2. The column of y-coefficients {{{matrix(3,1,1,-1,2)}}}
 
3. The column of z-coefficients {{{matrix(3,1,1,1,-3)}}} 
 
4. The column of constants:     {{{red(matrix(3,1,6,3,-4))}}}

I colored the column of constants red because that is
the column that "moves" from left to right in the 
determinants.
 
There are four determinants:
 
1. The determinant {{{D}}} consists of just the three columns
of x, y, and z coefficients. in that order, but does not
contain the column of constants.
 
{{{D=abs(matrix(3,3,1,1,1,2,-1,1,1,2,-3))}}}. 
 
It has value {{{D=13}}}.  I'm assuming you know how to find the
value of a 3x3 determinant, for that's a subject all by itself.
If you don't know how, ask me in the thank-you note form below
this problem.  I don't charge any money.  I do this for fun! 
 
2. The determinant {{{D[x]}}} is like the determinant {{{D}}}
except that the column of x-coefficients is replaced by the
column of constants.  {{{D[x]}}} does not contain the column 
of x-coefficients.
 
{{{D[x]=abs(matrix(3,3,red(6),1,1,red(3),-1,1,red(-4),2,-3))}}}.
 
It has value {{{D[x]=13}}}.
 
3. The determinant {{{D[y]}}} is like the determinant {{{D}}}
except that the column of y-coefficients is replaced by the
column of constants.  {{{D[y]}}} does not contain the column 
of y-coefficients.
 
{{{D[y]=abs(matrix(3,3,1,red(6),1,2,red(3),1,1,red(-4),-3))}}}.
 
It has value {{{D[y]=26}}}.
 
4. The determinant {{{D[z]}}} is like the determinant {{{D}}}
except that the column of z-coefficients is replaced by the
column of constants.  {{{D[z]}}} does not contain the column 
of z-coefficients.
 
{{{D[z]=abs(matrix(3,3,1,1,red(6),2,-1,red(3),1,2,red(-4)))}}}.
 
It has value {{{D[x]=39}}}.
 
Now the formulas for x, y and z are
 
{{{x=D[x]/D=13/13=1}}}
{{{y=D[y]/D=26/13=2}}}
{{{x=D[z]/D=39/13=3}}}

Now do the same with yours.  

[Hint as a check: Two of the values of your solution are
the same as two of the values in the problem I worked above.] 

Edwin</pre>