Can somebody solve this out? Identify Two additional vValues for x and y in a direct variation relationship where y=11 when x=8.

You can multiply both values, x and y by the same value and it still holds true.

y=22 when x=16

y=44 when x=32

y=55 when x=40

Direct variation: y=kx

Ask ourselves: what is k in this particular relationship? Substitute:

y=kx

11=8k

k=8/11

Two additional values for x and y; we can just throw in random numbers for x and see what y becomes. I'm going to use 11 and 22.

y = (8/11) x

y = (8/11) 11

y = 8

(11,8)

y = (8/11) x

y = (8/11) 22

y = (8/11) 2*11

y = 8*2

y = 16

(22,16)