You have
#[(-2,-3,|-color(white)(0)7),(5,-6,|color(white)(-)16)]#
After a quick switch of the last equation to a #ax + by = c# format, now we want
#[(1,0,|h),(0,1,|q)]#
So we know #x = h# and #y = q# as solutions.
Adding #5/2L_1# to #L_2# we have
#[(-2,-3,|-color(white)(0)7color(white)(+5/2(-7))),(5+5/2(-2),-6+5/2(-3),|color(white)(-)16+5/2(-7))]#
#[(-2,-3,|-color(white)(0)7),(0,-27/2,|color(white)()-3/2)]#
Multiplying #L_2# by #2#
#[(-2,-3,|color(white)(0)-7),(0,-27,|color(white)(0)-3)]#
Add #-1/9L_2# to #L_1#
#[(-2-0,-3+3,|color(white)(0)-7+1/3),(0,-27,|color(white)(0)-3)]#
#[(-2,0,|color(white)(0)-20/3),(0,-27,|color(white)(0)-3)]#
Multiply #L_1# by #-1/2#
#[(1,0,|color(white)(0)10/3),(0,-27,|color(white)(0)-3)]#
Multiply #L_2# by #-1/27#
#[(1,0,|color(white)(0)10/3),(0,1,|color(white)(0)1/9)]#
So #x = 10/3# and #y = 1/9#