For what value(s) of k will the relation not be a function?

A = {(1.5k−4, 7), (−0.5k+8, 15)}

for every value of x we want one and only one value of y

if
1.5 k - 4 = -0.5 k + 8
then we have TWO values of y for ONE value of x
k = 6 does it

You are welcome.

LOLOLLOL UR MOM

To determine whether the relation is a function, we need to check if each input value is associated with exactly one output value.

In this case, the relation is represented by set A = {(1.5k−4, 7), (−0.5k+8, 15)}. The x-values in the ordered pairs represent the input values, and the y-values represent the output values.

For the relation to be a function, each input value should correspond to only one output value. This means that if two different inputs have the same output, the relation will not be a function.

Let's equate the y-values in the ordered pairs and solve for k:

For the first ordered pair: 7 = 7
For the second ordered pair: 15 = 15

Since both equations are true, it means that regardless of the value of k, the y-values are the same for both ordered pairs.

Now, let's equate the x-values in the ordered pairs and solve for k:

For the first ordered pair: 1.5k−4 = −0.5k+8
For the second ordered pair: −0.5k+8 = 1.5k−4

Solving these equations will give us a value(s) of k where the ordered pairs have the same x-value. If there is more than one solution, then the relation will not be a function.

Let's solve the equations:

For the first ordered pair:
1.5k−4 = −0.5k+8
2k = 12
k = 6

For the second ordered pair:
−0.5k+8 = 1.5k−4
2k = 12
k = 6

Both equations have the same solution, k = 6. This means that for k = 6, the relation will not be a function because the same input value results in different output values.

Therefore, the relation is not a function for k = 6.

Thanks:)