Since equations are consistent, we find values of #x# and #y# first and then substitute them in the equation to find value of #k#.
#x+3y+2 = 0# -------> equation 1
#4y+2x=k# ----------> equation 2
#x-2y=3# ------------> equation 3
From equation 1; make #x# the subject.
#x-2y=3#
#color(red)(x=3+2y)#
Substitute #x=3+2y# in equation 1
#x+3y+2=0#
#color(red)((3+2y))+3y+2=0#
#3+2y+3y+2=0#
#3+5y+2=0#
#5y=-2-3#
#5y=-5#
#color(red)(y=-1)#
Now, substitute value of #y=-1# in equation 3 to get value of #x#
#x-2y=3#
#x-2(-1)=3#
#x+2=3#
#x=3-2#
#color(red)(x=1)#
Check the answer of values of #x# and #y# before finding value of #k#
#x+3y+2=0#
#1+3(-1)+2=0#
#1-3+2=0#
#-2+2=0# ------> so values of #x# and #y# are correct.
Final step is to substitute values of #x# and #y# in equation 2 to find value of #k#:
#4y+2x=k#
#4(-1) + 2(1)=k#
#-4+2=k#
#-2=k#
Therefore, #color(red)(k = -2)#
