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Math helpppppp algerbraa 2

Math helpppppp algerbraa 2-example-1

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Answers:

Option 1.) -4|6x + 3| < 16

Option 2.) 8|6x + 3|+ 13 > 5

Option 4.) |4x - 9| ≥ -12

Step-by-step explanation:

All real numbers (R) contain rational and irrational numbers. Therefore, the correct answers are:

Option 1.) -4|6x + 3| < 16

Multiply both sides of the inequality by (-1) to reverse the inequality symbol:

(-1 ) (-4 |6x + 3| ) < 16 (-1 )

4|6x + 3| > - 16

Divide both sides of the inequality by 4:


(6|6x+3|)/(4) > (-16)/(4)

|6x + 3| > - 4 → true for all x values ( – ∞, ∞)

Step-by-step explanation: If the absolute value is greater than a negative number, the solution is all real numbers.

Option 2.) 8|6x + 3|+ 13 > 5

Subtract 13 from both sides of the inequality:

8|6x + 3|+ 13 - 13 > 5 - 13

8|6x + 3| > - 8

Divide both sides of the inequality by 8:


(8|6x+3|)/(8) > (-8)/(8)

|6x + 3| > - 1 → true for all x values ( – ∞, ∞)

Step-by-step explanation: If the absolute value is greater than a negative number, the solution is all real numbers.

Option 4.) |4x - 9| ≥ -12 → true for all x values ( – ∞, ∞)

Explanation: If the absolute value is greater than or equal to a negative number, the solution is all real numbers. The absolute value of something will always be greater than a negative number.

Example: if x = - 50, then:

|4(-50) - 9| ≥ -12

|-200 - 9| ≥ -12

|-209| ≥ -12

209 ≥ -12 is a true statement. Therefore, x = - 50 is a solution.

If x = 0, then:

|4(0) - 9| ≥ -12

|0 - 9| ≥ -12

|- 9| ≥ -12

9 ≥ -12 is a true statement. Therefore, x = 0 is also a solution.

This proves that regardless of any x values, as long as the inequality is either greater than/ greater than or equal to a negative number, all real numbers are valid solutions for x because the absolute value of something will always be greater than a negative number.

User Nikunj Kumbhani
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