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

Answer 1

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

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

2. Divide both sides by 2: \[ x \leq 6 \]

Now, any value of \( x \) that is less than or equal to 6 will satisfy the inequality.

Here are two example values for \( x \):

• \( x = 0 \)
• \( x = 5 \)

Both values satisfy the inequality \( 8 + 2x \leq 20 \):

• For \( x = 0 \): \[ 8 + 2(0) = 8 \leq 20 \quad \text{(True)} \]
• For \( x = 5 \): \[ 8 + 2(5) = 18 \leq 20 \quad \text{(True)} \]

Thus, two valid values for \( x \) are 0 and 5.

Answer 2

another

Answer 3

Sure! We can choose different values for \( x \) that still satisfy the inequality \( 8 + 2x \leq 20 \) (or \( x \leq 6 \)).

Here are two more example values:

• \( x = 3 \)
• \( x = 6 \)

Let's check both:

1. For \( x = 3 \): \[ 8 + 2(3) = 8 + 6 = 14 \leq 20 \quad \text{(True)} \]

2. For \( x = 6 \): \[ 8 + 2(6) = 8 + 12 = 20 \leq 20 \quad \text{(True)} \]

Both values \( 3 \) and \( 6 \) satisfy the inequality \( 8 + 2x \leq 20 \).

So, two additional valid values for \( x \) are 3 and 6.

Answer 4

another

Answer 5

Certainly! Here are two more values for \( x \) that satisfy the inequality \( 8 + 2x \leq 20 \) (or \( x \leq 6 \)):

• \( x = 1 \)
• \( x = 4 \)

Let's verify both:

1. For \( x = 1 \): \[ 8 + 2(1) = 8 + 2 = 10 \leq 20 \quad \text{(True)} \]

2. For \( x = 4 \): \[ 8 + 2(4) = 8 + 8 = 16 \leq 20 \quad \text{(True)} \]

Both values \( 1 \) and \( 4 \) satisfy the inequality \( 8 + 2x \leq 20 \).

So, two additional valid values for \( x \) are 1 and 4.

Answer 6

another includeing 10 8 or 6 4

Answer 7

To clarify, the inequality \(8 + 2x \leq 20\) simplifies to \(x \leq 6\). Thus, any values of \(x\) that are less than or equal to 6 will satisfy the inequality.

Given that, we can include the values 4 and 6 (which you mentioned) as well as others:

1. \( x = 6 \): \[ 8 + 2(6) = 8 + 12 = 20 \leq 20 \quad \text{(True)} \]

2. \( x = 4 \): \[ 8 + 2(4) = 8 + 8 = 16 \leq 20 \quad \text{(True)} \]

So, \(4\) and \(6\) are two values that satisfy the inequality.

For additional values without violating the constraint, consider:

3. \( x = 0 \): \[ 8 + 2(0) = 8 \leq 20 \quad \text{(True)} \]

4. \( x = 2 \): \[ 8 + 2(2) = 8 + 4 = 12 \leq 20 \quad \text{(True)} \]

To summarize:

• Values 4 and 6 satisfy \(8 + 2x \leq 20\).
• Additional values could be 0 and 2.