We can use Gauss Jordan Elimindation and an augmented matrix to solve this system. For convenience sake, I will rearrange the equations slightly. In referring to a row, I'll use the designation #R_n# to refer to Row #n.
We begin with the augmented matrix:
#[ (-1,1,3,2,|,4),(5,-6,3,3,|,2),(4,-5,6,5,|,6),(-3,3,9,6,|,12) ] {:((-1)*R_1),(->),(),() :}#
#[ (1,-1,-3,-2,|,-4),(5,-6,3,3,|,2),(4,-5,6,5,|,6),(-3,3,9,6,|,12) ] {:(->),(-5R_1),(-4R_1),(+3R_1) :}#
#[ (1,-1,-3,-2,|,-4),(0,-1,18,13,|,22),(0,-1,18,13,|,22),(0,0,0,0,|,0) ] {:(->),((-1)*R_2),(-R_2),() :}#
#[ (1,-1,-3,-2,|,-4),(0,1,-18,-13,|,-22),(0,0,0,0,|,0),(0,0,0,0,|,0) ] {:(+R_2),(->),(),() :}#
#[ (1,0,-21,-15,|,-26),(0,1,-18,-13,|,-22),(0,0,0,0,|,0),(0,0,0,0,|,0) ] #
This is as far as we can get. The bottom two rows being completely zeroed out indicates that the last two variables #x_3# and #x_4# are not specified by the original system of equations. Their values can be any values we wish. The top two equations translate to the following:
#{(x_1-21x_3-15x_4 = -26),(x_2-18x_3-13x_4=-22) :}#
We can rearrange these two equations to solve for #x_1# and #x_2#, and combine that with #x_3# and #x_4# for a complete solution set:
#{(x_1 = 21x_3+15x_4-26),(x_2=18x_3+13x_4-22),(x_3),(x_4) :}#
