The correct option is B concurrent
y=5x−3 ....eq1
y=4−2x ....eq2
Put y=4−2x in eq1
4−2x=5x−3
−7x=−7
x=1
Put x=1 in eq1 then
y=5×1−3
y=2
Put x=1 and y=2 in 2x+3y=8
L.H.S=2×1+3×2=2+6=8=R.H.S
∴all these lines whose equation are y=5x−3,y=4−2y,2x+3y=8 passes through (1,2).
∴ they are concurrent lines and their common point of intersection is (1,2).