Question 1106479
<pre><font size = 4 color = "indigo"><b>
2x+y+2z=4
2x+2y  =5
2x- y+6z=2

Write it this way, putting in 1 coefficients and 0z
fo the missing z-term in the second equation.

2x + 1y + 2z = 4
2x + 2y + 0z = 5
2x - 1y + 6z = 2

Then do it step by step just like this one:
-----------------------------

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

Write it this way:

1x + 1y + 1z =  6
2x - 1y + 1z =  3
1x + 2y - 3z = -4

{{{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))}}}
 
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 how in the note form below and I'll
get back to you by email.  No charge. 
 
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}}}
 
Edwin</pre>