Question

Which of the following ordered pairs is not a solution to the inequality y greater than or equal to -4x - 2?


(-1, 0)

(0, 0)

(4, 4)

(1, 3)

Answer 1

When you do this type of problem, just remember that the first number is the x-coordinate, and the second number is the y-coordinate.
For each number, you're going to see if the inequality is true or false when you make a substitution. Let's take (-1, 0) as an example.
\[y \ge -4x-2\]\[0 \ge -4(-1)-2\]\[0 \ge 4-2\]\[0 \ge 2\]but zero isn't larger than 2, so this is false. This means that (-1, 0) doesn't work.

You have to go through and check each of those 4. There could be multiple right answers.

Answer 2

so are you saying (1,0) is the correct answer ?

Answer 3

I'm saying that (-1, 0) is not correct.

Answer 4

ok. can you help me with another one please?

Answer 5

@jabberwock

Answer 6

Sure thing.

Answer 7

Which of the following ordered pairs is a solution to the inequality y greater than or equal to 4x - 2?

(0, -4)

(0, 0)

(0, -3)

(0, -5)

Answer 8

You're going to do this one the same way. Did you finish the last one? (You have to check all 4 because there might be multiple correct answers)

Answer 9

i dont understand how to check them

Answer 10

Let's look back at the example of (-1, 0). -1 is an x-coordinate, and 0 is a y-coordinate, right?

Answer 11

yes

Answer 12

Okay. So now you're going to take your x-coordinate and your y-coordinate, and you're going to plug them into your inequality.
\[y \ge -4x-2\]\[0 \ge -4(-1) -2\]
Okay with that part?

Answer 13

yes

Answer 14

Alright, so if we do the arithmetic and simplify it all down, we get
\[0 \ge 2\]is that a true statement or a false statement?

Answer 15

false

Answer 16

Right. Now, since it's false, we know it doesn't work. If we would have gotten a true statement (something like 17>=0), then we would have said "yes." Make sense?

Answer 17

yes

Answer 18

Okidoki. Sorry to do this, but I have to get going. Maybe try reposting one of the questions if you get stuck. Good luck!