Question 135991
When you say 'coordinate' method, do you mean to solve the system by graphing?  I certainly hope not because there is no way to render a three-dimensional coordinate system on this site.  However, I'll do one of your systems by the other three methods.


1. {{{x+y+z=6}}}
2. {{{2x-y+3z=9}}}
3. {{{-x+2y+z=6}}}


Elimination:
Replace Eq. 2 with the sum of Eq. 2 and two times Eq. 3.
Replace Eq. 3 with the sum of Eq. 1 and Eq. 3.
1. {{{x+y+z=6}}}
2. {{{0x+3y+5z=21}}}
3. {{{0x+3y+2z=12}}}


Replace Eq. 3 with the sum of Eq. 2 and -1 times Eq. 3.
1. {{{x+y+z=6}}}
2. {{{0x+3y+5z=21}}}
3. {{{0x+0y+3z=9}}}


Divide Eq 3 by 3
1. {{{x+y+z=6}}}
2. {{{0x+3y+5z=21}}}
3. {{{0x+0y+z=3}}}


Replace Eq. 1 with the sum of Eq. 1 and -1 times Eq. 3
Replace Eq. 2 with the sum of Eq. 2 and -5 times Eq. 3
1. {{{x+y+0z=3}}}
2. {{{0x+3y+0z=6}}}
3. {{{0x+0y+z=3}}}


Divide Eq. 2 by 3
1. {{{x+y+0z=3}}}
2. {{{0x+y+0z=2}}}
3. {{{0x+0y+z=3}}}


Replace Eq. 1 with the sum of Eq. 1 and -1 times Eq. 2
1. {{{x+0y+0z=1}}}
2. {{{0x+y+0z=2}}}
3. {{{0x+0y+z=3}}}


Eliminate the variables with zero coefficients
1. {{{x=1}}}
2. {{{y=2}}}
3. {{{z=3}}}


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Substitution:

1. {{{x+y+z=6}}}
2. {{{2x-y+3z=9}}}
3. {{{-x+2y+z=6}}}


Solve Eq. 1 for x:
{{{x=6-y-z}}}


Substitute the expression for x into the other two equations
2. {{{2(6-y-z)-y+3z=9}}}
3. {{{-(6-y-z)+2y+z=6}}}


Distribute and collect like terms:
2. {{{12-2y-2z-y+3z=9}}} => {{{-3y+z=-3}}} (New Eq. 2)

3. {{{-6+y+z+2y+z=6}}} => {{{3y+2z=12}}} (New Eq. 3)


Solve equation 2 for y:
{{{-3y=-3-z}}}
{{{y=1+(z/3)}}}


Substitute this expression for y into the new Eq. 3
{{{3(1+z/3)+2z=12}}}


Distribute, simplify, collect terms
{{{3+3z/3+2z=12}}}
{{{z+2z=12-3}}}
{{{3z=9}}}
{{{z=3}}}


Substitute this value for z into the new Eq. 2
{{{-3y+3=-3}}}
{{{-3y=-6}}}
{{{y=2}}}


Substitute the values for y and z into the original equation 1:
{{{x+y+z=6}}}
{{{x+2+3=6}}}
{{{x=1}}}


============================================================
Cramer's Rule

Create and evaluate the coefficient determinant


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


If you don't remember how to evaluate a determinant, here is a little cartoon that shows the process.  You have to take your coefficient determinant and make a 5X3 matrix by repeating the 1st and 2nd columns first.  Then follow the process shown:


{{{cartoon(
(matrix(3,5,red(a),b,c,a,b,d,red(e),f,d,e,g,h,red(i),g,h))=red(aei),
(matrix(3,5,a,red(b),c,a,b,d,e,red(f),d,e,g,h,i,red(g),h))=aei+red(bfg),
(matrix(3,5,a,b,red(c),a,b,d,e,f,red(d),e,g,h,i,g,red(h)))=aei+bfg+red(cdh),(matrix(3,5,a,b,red(c),a,b,d,red(e),f,d,e,red(g),h,i,g,h))=aei+bfg+cdh-(red(gec)),
(matrix(3,5,a,b,c,red(a),b,d,e,red(f),d,e,g,red(h),i,g,h))=aei+bfg+cdh-(gec+red(hfa)),
(matrix(3,5,a,b,c,a,red(b)b,d,e,f,red(d),e,g,h,red(i),g,h))=aei+bfg+cdh-(gec+hfa+red(idb)))}}}


Yours evaluates to: -9, so we say {{{D=-9}}}


Next, replace the first column representing the coefficients on the x terms with the constant term values.


{{{D[x]=(matrix(3,3,6,1,1,9,-1,3,6,2,1))}}}, and do the determinant evaluation on t{{{D[x]}}}


This works out to -9 as well, so you can say {{{D[x]=-9}}}


Cramer's rule says:

{{{x=D[x]/D}}}.  In this case:  {{{-9/-9=1}}} (Just as I most sincerely hope you expected)


As you might suspect, you replace the second column with the coefficients to get {{{D[y]}}} and the third column to get {{{D[z]}}}.


Then {{{y=D[y]/D}}} and {{{z=D[z]/D}}}


I'll give you the {{{D[y]}}} and {{{D[z]}}} matrices and you can do the calculations for yourself.


{{{D[y]=(matrix(3,3,1,6,1,2,9,3,-1,6,1))}}}


{{{D[z]=(matrix(3,3,1,1,6,2,-1,9,-1,2,6))}}}