From the first equation, write y in terms of x

y = x – 5 …….(i)

Substitute different values of x in (i) to get different values of y

For x = 0, y = 0 – 5 = -5

For x = 2, y = 2 – 5 = -3

For x = 5, y = 5 – 5 = 0

Thus, the table for the first equation (x – y – 5 = 0) is

Now, plot the points A(0, -5), B(2, -3) and C(5, 0) on a graph paper and join A, B and C to get the graph of x – y – 5 = 0.

From the second equation, write y in terms of x

y = \(\frac{15-3x}5\)……(ii)

Now, substitute different values of x in (ii) to get different values of y

For x = -5, y = \(\frac{15+15}5\) = 6

For x = 0, y = \(\frac{15-0}5\) = 3

For x = 5, y = \(\frac{15-15}5\) = 0

So, the table for the second equation (3x + 5y – 15 = 0) is

Now, plot the points D(-5, 6), E(0, 3) and F(5, 0) on the same graph paper and join D, E and F to get the graph of 3x + 5y – 15 = 0

From the graph, it is clear that, the given lines intersect at (5, 0).

So, the solution of the given system of equation is (5, 0).

From the graph, the vertices of the triangle formed by the given lines and the y-axis are 0, 3), (0, -5) and (5, 0).

Now, draw a perpendicular from the intersection point C on the y-axis. So,

Area (∆CEA) = \(\frac{1}2\) × EA × OC

= \(\frac{1}2\) × 8 × 5

= 20 sq. units

Hence, the vertices of the triangle formed by the given lines and the y-axis are (0, 3), (0, -5) and (5, 0) and the area of the triangle is 20 sq. units.