Question 959344
Given the following equations of a line: 
X-5y=17
Y-2=6(x+2)
Y=(7/6x)+(14/3) 
Find the vertices of the triangle
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vertices of the triangle are the points of intersection of the 3 equations above
x-5y=17
5y=x-17(eq.1)
..
y-2=6x+12
y=6x+14(eq.2)
..
y=7x/6+4/3
6y=7x+8(eq.3)
..
5y=x-17(eq. 1)
5y=30x+70(eq.2*5)
subtract:
-29x-87=0
29x=-87
x=-87/29=-3
5y=x-17=-20
y=-4
point of intersection: (-3,-4)
..
5y=x-17(eq. 1)
6y=7x+8(eq.3)
..
30y=6x-102
30y=35x+40
subtract
-29x-142=0
29x=-142
x=-4.90
5y=x-17=-21.9
y=-4.38
point of intersection: (-4.90,-4.38)
..
y=6x+14(eq.2)
6y=7x+8(eq.3)
..
6y=36x+84
6y=7x+8
subtract:
29x+76=0
29x=-76
x=-2.62
y=6x+14=-1.72
point of intersection: (-2.62,-1.72)
..
vertices of the triangle are: (-3,-4),(-4.90,-4.38), (-2.62, -1.72)