Show that the lines 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

When lines are concurrent they intersect at same point.

Solving, 2x + 5y = 1 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 6 + 3y

Substituting x = 3y + 6 in 2x + 5y = 1 we get,

⇒ 2(3y + 6) + 5y = 1

⇒ 6y + 12 + 5y = 1

⇒ 11y = -11

⇒ y = -1.

Substituting y = -1 in x = 6 + 3y we get,

x = 6 + 3(-1) = 6 - 3 = 3.

∴ 2x + 5y = 1 and x - 3y = 6 intersect in the point (3, -1).

Solving, x + 5y + 2 = 0 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 3y + 6

Substituting x = 3y + 6 in x + 5y + 2 = 0 we get,

⇒ 3y + 6 + 5y + 2 = 0

⇒ 8y + 8 = 0

⇒ 8y = -8

⇒ y = -1.

Substituting y = -1 in x = 3y + 6 we get,

x = 3(-1) + 6 = -3 + 6 = 3.

∴ x + 5y + 2 = 0 and x - 3y = 6 intersect in the point (3, -1).

Hence, proved that 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.