Triangles JFE and JLK are similar provided that line segments FE and LK are parallel. This means that corresponding parts of these triangles occur proportionally to one another. In particular,

From this relation we get

Now we can solve for
:




Step-by-step explanation: First, let's graph this down.
With all the points touching except (-2, -6) and (8, 18), it's safe to say that these are the coordinates that have to be removed. Also, just for an extra help, the linear line is 2x + 1 = y
Answer:
6x−11y+15
Step-by-step explanation:
Distribute the Negative Sign:
=5x−6y+7+−1(−x+5y−8)
=5x+−6y+7+−1(−x)+−1(5y)+(−1)(−8)
=5x+−6y+7+x+−5y+8
Combine Like Terms:
=5x+−6y+7+x+−5y+8
=(5x+x)+(−6y+−5y)+(7+8)
=6x+−11y+15
Answer:
6x−11y+15
I hope I helped!!
Answer:
so if x = 0, x = 2, and x = 4
3(0) + 2 = 0 + 2 = 2, so when x = 0, y = 2
3(2) + 2 = 6 + 2 = 8, so when x = 2, y = 8
3(4) + 2 = 12 + 2 = 14, so when x = 4, y = 14