Which ordered pairs make the open sentence true?

3x+y<14

Select all the correct answers.

(3, 6)
left parenthesis 3 comma 6 right parenthesis

(7, −7)
left parenthesis 7 comma negative 7 right parenthesis

(6, 0)
left parenthesis 6 comma 0 right parenthesis

(−1, 6)
left parenthesis negative 1 comma 6 right parenthesis

(4, −1)

An inequality can be defined as a mathematical relation that compares two (2) or more integers and variables in an algebraic expression based on any of the following arguments:

Less than (<).
Greater than (>).
Less than or equal to (≤).
Greater than or equal to (≥).

For this exercise, you should evaluate each of the given expressions to determine which inequalities is true when the ordered pairs are substituted. This ultimately implies that, you would have to substitute the ordered pairs into each of the given algebraic expressions and then evaluate.

When ordered pairs = (3, 6), we have:

3x + y < 14

3(3) + 6 < 14

9 + 6 < 14

15 < 14 (False).

When ordered pairs = (7, -7), we have:

3x + y < 14

3(7) + (-7) < 14

21 - 7 < 14

14 < 14 (False).

When ordered pairs = (6, 0), we have:

3x + y < 14

3(6) + 0 < 14

18 - 0 < 14

18 < 14 (False).

When ordered pairs = (-1, 6), we have:

3x + y < 14

3(-1) + 6 < 14

-3 + 6 < 14

3 < 14 (True).

When ordered pairs = (4, -1), we have:

3x + y < 14

3(4) + (-1) < 14

12 - 1 < 14

11 < 14 (True).

Read more on ordered pairs here: https://brainly.com/question/22683073

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