Solution:

Given,

2y – x = 8……. (i)

5y – x = 14……. (ii)

y – 2x = 1……… (iii)

For equation (i),

⇒ y = (x + 8)/2

When x = -4, we have y = (-4 + 8)/2 = 2

When x = 0, we have y = (0 + 8)/2 = 4

Thusm we have the following table giving points on the line 2y – x = 8

x	-4	0
y	2	4

For equation (ii),

We solve for y:

⇒ y = (x + 14)/5

So, when x = -4

y = ((-4) + 14)/5= 2

And, when x = 1

⇒ y = (1 + 14)/5= 3

Thusm we have the following table giving points on the line 5y – x = 14

x	-4	1
y	2	3

Finally, for equation (iii),

⇒ y = (2x + 1)

When x = -1, we have y = (2(-1) + 1) = -1

When x = 1, we have y = (2(1) + 1) = 3

Thusm we have the following table giving points on the line y – 2x = 1

x	-1	1
y	1	3

Graph of the equations (i), (ii) and (iii) is as below:

From the above graph, we observe that the lines taken in pairs intersect at points A(-4,2), B(1,3) and C(2,5).

Hencem the vertices of the triangle are A(-4, 2), B(1, 3) and C(2,5).