Question 1004053
5x-6y+4z=15,
7x+4y-3z=19,
2x+y+6z=46
solve using cramers rule
<pre> 
Write in the 1 coefficient:

{{{system(5x-6y+4z=red(15),
7x+4y-3z=red(19),
2x+1y-6z=red(46))}}}

Cramer's rule:
 
There are 4 columns,
 
1. The column of x-coefficients {{{matrix(3,1,5,7,2)}}}
 
2. The column of y-coefficients {{{matrix(3,1,-6,4,1)}}}
 
3. The column of z-coefficients {{{matrix(3,1,4,-3,6)}}} 
 
4. The column of constants:     {{{red(matrix(3,1,15,19,46))}}}
 
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(3,3,5,-6,4,7,4,-3,2,1,6))}}}. 
 
It has value {{{D=419}}}.  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, post again asking how. 
 
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(15),-6,4,red(19),4,-3,red(46),1,6))}}}.
 
It has value {{{D[x]=1257}}}.
 
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(3,3,5,red(15),4,7,red(19),-3,2,red(46),6))}}}.
 
It has value {{{D[y]=1676}}}.
 
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(3,3,5,-6,red(15),7,4,red(19),2,1,red(46)))}}}.
 
It has value {{{D[z]=2514}}}.
 
Now the formulas for x, y and z are
 
{{{x=D[x]/D=1257/419=3}}}
{{{y=D[y]/D=1676/419=4}}}
{{{x=D[z]/D=2514/419=6}}}
 
Edwin</pre>