Question 19634
SEE THE FOLLOWING AND COME BACK IF YOU HAVE DIFFICULTY.HERE C,CX,CY,CZ REFER TO YOUR D,DX,DY,DZ...JUST A DIFFERENCE IN NOMENCLATURE.I SHOWED IN DETAIL A 2X2 DETERMINANT AND THEN IN BRIEF A 3X3 DETERMINANT

 2x+y=4 
3x-y=6 
make a deteminant with coefficients of x (2,3)and y(1,-1) in the 2 eqns.call it C.(Actually for a determinant as you know ,the numbers are contained in vertical bars at either end like |xx|,but in the following the bars are omitted due to difficulty in depiction.you may assume the bars are present)
C={{{matrix(2,2,2,1,3,-1)}}}=2*(-1)-(1*3)=-5
..now use the constants (4,6)to replace coefficients of x(2,3) in the above determinant C...call it CX..
CX={{{matrix(2,2,4,1,6,-1)}}}=4*(-1)-1*6=-4-6=-10
..now use the constants (4,6)to replace coefficients of y(1,-1) in the above determinant C...call it CY..
CY={{{matrix(2,2,2,4,3,6)}}}=2*6-3*4=12=12=0
..now cramers rule says that
{{{(x/CX)=(y/CY)=(1/C)}}}..so we get
x/(-10)=y/0=1/-5
x=-10/-5=10/5=2
y=0/-5=0
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so using the above method you can do the next problem ..here due to presence of 3 variables you will get 3rd.order determinants...4 in all...namely C,CX,CY and CZ,the last formula also extends to include z ,
{{{(x/CX)=(y/CY)=(z/CZ)=(1/C)}}}..
 but the procedure is same ..

 2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7 ...
...just to give you the idea 
C={{{matrix(3,3,2,3,1,1,1,-2,-3,0,1)}}}..and 
CZ={{{matrix(3,3,2,3,5,1,1,-2,-3,0,7)}}}..etc..hope you can work out the rest