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What are two values for `x` that would make `8+2x\le20` true.

3 answers

To solve the inequality \( 8 + 2x \le 20 \), we first need to isolate \( x \).

  1. Subtract 8 from both sides: \[ 2x \le 20 - 8 \] \[ 2x \le 12 \]

  2. Now, divide both sides by 2: \[ x \le \frac{12}{2} \] \[ x \le 6 \]

This means that \( x \) can take any value less than or equal to 6.

Here are two values for \( x \) that satisfy the inequality:

  1. \( x = 6 \)
  2. \( x = 5 \)

Both values fulfill the condition \( 8 + 2x \le 20 \):

  • For \( x = 6 \): \[ 8 + 2(6) = 8 + 12 = 20 \quad (\text{valid since } 20 \le 20) \]

  • For \( x = 5 \): \[ 8 + 2(5) = 8 + 10 = 18 \quad (\text{valid since } 18 \le 20) \]

Thus, \( x = 6 \) and \( x = 5 \) are two valid solutions.

Are you sure its not 6 and 4?

Yes, I can confirm that \( x = 6 \) and \( x = 5 \) are both valid solutions based on the inequality \( 8 + 2x \le 20 \).

To clarify further, any value of \( x \) that is less than or equal to 6 will satisfy the inequality because we derived that \( x \le 6 \).

However, \( x = 4 \) is also a valid solution because:

  • For \( x = 4 \): \[ 8 + 2(4) = 8 + 8 = 16 \quad (\text{valid since } 16 \le 20) \]

Thus, both pairs of values \( (6, 5) \) and \( (6, 4) \) are valid solutions to the inequality, along with any value less than or equal to 6.

So, the correct response could include \( 6 \) and either \( 5 \) or \( 4 \) as valid values, among many others such as \( 0 \), \( -1 \), etc. Thank you for pointing that out!