Determine, algebraically, the vertices of the triangle formed by the lines. 3x - y = 32x - 3y = 2x + 2y = 8

To obtain the vertices of triangle formed by above equations : 3x - y = 3, 2x - 3y = 2 and x + 2y = 8.

First we will solve equations:
3x - y = 3 and 2x - 3y = 2
then 2x - 3y = 2 and x + 2y = 8
and lastly x + 2y = 8 and 3x - y = 3

Now on solving equations 3x - y = 3 and 2x - 3y = 2, we get x = 1 and y = 0,

So one of the vertex is (1, 0)

Similarly, on solving pair of equations 2x - 3y = 2 and x + 2y = 8 we get x = 4 and y = 2 so second vertex will be (4, 2)

On solving x + 2y = 8 and 3x - y = 3 we get x = 2 and y = 3.Hence third vertex is (2, 3).So,the vertices of Triangle are (1, 0),(4, 2) and (2, 3)

The first one is 2/3the second is 1/160the third is 4/125