For each value of x, determine whether it is a solution to -5+6x≤37. ? 00 Is it a solution? Yes No 0 10 -3 7

Answer 1

To determine whether each value of \( x \) is a solution to the inequality \( -5 + 6x \leq 37 \), we can solve the inequality for \( x \) first.

1. Start with the inequality: \[ -5 + 6x \leq 37 \]

2. Add 5 to both sides: \[ 6x \leq 37 + 5 \] \[ 6x \leq 42 \]

3. Divide both sides by 6: \[ x \leq 7 \]

Now we can check each value of \( x \):

1. For \( x = 0 \): \[ 6(0) = 0 \quad \Rightarrow \quad -5 + 0 = -5 \leq 37 \] Yes, it is a solution.

2. For \( x = 10 \): \[ 6(10) = 60 \quad \Rightarrow \quad -5 + 60 = 55 \not\leq 37 \] No, it is not a solution.

3. For \( x = -3 \): \[ 6(-3) = -18 \quad \Rightarrow \quad -5 - 18 = -23 \leq 37 \] Yes, it is a solution.

4. For \( x = 7 \): \[ 6(7) = 42 \quad \Rightarrow \quad -5 + 42 = 37 \leq 37 \] Yes, it is a solution.

So, summarizing the results:

• \( x = 0 \): Yes
• \( x = 10 \): No
• \( x = -3 \): Yes
• \( x = 7 \): Yes

Answer 2

one of them are in corect and its not 0

Answer 3

Let's go through the values one by one again to ensure we evaluate them correctly against the inequality \( -5 + 6x \leq 37 \).

Given inequality:

\[ -5 + 6x \leq 37 \]

Solving the inequality:

1. Add 5 to both sides: \[ 6x \leq 42 \]

2. Divide both sides by 6: \[ x \leq 7 \]

Now, let's check each value:

1. For \( x = 0 \): \[ -5 + 6(0) = -5 \quad \Rightarrow \quad -5 \leq 37 \quad \text{(True)} \] Conclusion: Yes, \( x = 0 \) is a solution.

2. For \( x = 10 \): \[ -5 + 6(10) = -5 + 60 = 55 \quad \Rightarrow \quad 55 \leq 37 \quad \text{(False)} \] Conclusion: No, \( x = 10 \) is not a solution.

3. For \( x = -3 \): \[ -5 + 6(-3) = -5 - 18 = -23 \quad \Rightarrow \quad -23 \leq 37 \quad \text{(True)} \] Conclusion: Yes, \( x = -3 \) is a solution.

4. For \( x = 7 \): \[ -5 + 6(7) = -5 + 42 = 37 \quad \Rightarrow \quad 37 \leq 37 \quad \text{(True)} \] Conclusion: Yes, \( x = 7 \) is a solution.

Summary of Correct Answers:

• \( x = 0 \): Yes
• \( x = 10 \): No
• \( x = -3 \): Yes
• \( x = 7 \): Yes

Thus, \( x = 0 \) was indeed a solution. If there's a specific error you suspect, please clarify, and we can re-evaluate that particular value!