Question

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

Answer 1

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)

Answer 2

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