For what value(s) of k will the relation not be a function? #1: A = {(k^2, 16), (4k, 32)}
#2: B = {(k^2-5k, 10), (k+7, 4)}
#3: R = {(k^3-5k^2+3k, -5), (-k, 4)}
#4: S = {(|k+1|+2, 4), (8, 7)}
Will give 100 points and Brainliest!

Step-by-step explanation:

The relation is not a function when x -value is same but y- values are different.

#1

#2

#3

#4

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1