If 4 , which of the following must be true? (5 (|x+3| > 2 \(-(\mathrm{x}+5)\) is positive

1. Translate the problem requirements: We need to solve the inequality \(4 < \frac{7-\mathrm{x}}{3}\) to find the range of x values, then determine which of the three given statements must always be true for any x in that range.
2. Solve the core inequality: Use algebraic manipulation to isolate x and determine the constraint on x values.
3. Evaluate each statement systematically: Test whether each of the three statements (I, II, III) must be true given our constraint on x.
4. Identify the combination of true statements: Match our findings to the answer choices to determine which statements must always hold.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to do. We have an inequality involving x, and we need to figure out which of three statements must always be true.

The inequality is: \(4 < \frac{7-\mathrm{x}}{3}\)

This tells us something about the possible values of x. Once we know what values x can take, we'll check each of the three statements:

• Statement I: \(5 < \mathrm{x}\) (meaning x is greater than 5)
• Statement II: \(|\mathrm{x}+3| > 2\) (meaning the distance from x to -3 is greater than 2)
• Statement III: \(-(\mathrm{x}+5)\) is positive (meaning the opposite of (x+5) is positive)

Our job is to find which of these statements must be true for ANY value of x that satisfies our original inequality.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Solve the core inequality

Now let's solve \(4 < \frac{7-\mathrm{x}}{3}\) to find what values x can have.

Starting with: \(4 < \frac{7-\mathrm{x}}{3}\)

To get rid of the fraction, let's multiply both sides by 3:
\(4 \times 3 < \frac{7-\mathrm{x}}{3} \times 3\)
\(12 < 7-\mathrm{x}\)

Now we need to isolate x. Let's subtract 7 from both sides:
\(12 - 7 < 7 - \mathrm{x} - 7\)
\(5 < -\mathrm{x}\)

To get x by itself, we multiply both sides by -1. Remember, when we multiply an inequality by a negative number, we must flip the inequality sign:
\(-5 > \mathrm{x}\)

This is the same as saying: \(\mathrm{x} < -5\)

So our constraint is that x must be less than -5.

Process Skill: MANIPULATE - Careful algebraic manipulation while preserving inequality relationships

3. Evaluate each statement systematically

Now we know that \(\mathrm{x} < -5\). Let's check each statement to see if it must always be true.

Statement I: \(5 < \mathrm{x}\)
This says x is greater than 5. But we found that \(\mathrm{x} < -5\), which means x is less than -5. Since -5 is much less than 5, there's no way x can be greater than 5. Statement I is FALSE.

Statement II: \(|\mathrm{x}+3| > 2\)
This asks about the absolute value of (x+3). Since \(\mathrm{x} < -5\), let's see what x+3 looks like:
If \(\mathrm{x} < -5\), then \(\mathrm{x}+3 < -5+3\), so \(\mathrm{x}+3 < -2\)

Since x+3 is always less than -2 (meaning it's negative and its distance from 0 is greater than 2), we have:
\(|\mathrm{x}+3| = -(\mathrm{x}+3) > -(-2) = 2\)

So \(|\mathrm{x}+3| > 2\) is always TRUE.

Statement III: \(-(\mathrm{x}+5)\) is positive
Since \(\mathrm{x} < -5\), we have \(\mathrm{x}+5 < -5+5\), so \(\mathrm{x}+5 < 0\)
This means x+5 is negative.
Therefore, \(-(\mathrm{x}+5)\) is the opposite of a negative number, which makes it positive.
Statement III is TRUE.

Process Skill: CONSIDER ALL CASES - Systematically checking each statement against our constraint

4. Identify the combination of true statements

From our analysis:

• Statement I is FALSE
• Statement II is TRUE
• Statement III is TRUE

So statements II and III must be true, while statement I is false.

Looking at our answer choices:

1. II only - No, because III is also true
2. III only - No, because II is also true
3. I and II only - No, because I is false and III is true
4. II and III only - Yes, this matches our findings
5. I, II and III - No, because I is false

Final Answer

The correct answer is D) II and III only.

Statements II and III must always be true when \(4 < \frac{7-\mathrm{x}}{3}\), while statement I is never true under this condition.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "must be true" means
Students often confuse "must be true" with "could be true." They might pick statements that are sometimes true for certain values of x, rather than statements that are ALWAYS true for ALL values of x that satisfy the constraint. This leads to incorrect evaluation of the Roman numeral statements.

2. Not recognizing the need to find the constraint on x first
Some students might try to evaluate each Roman numeral statement independently without first solving the inequality \(4 < \frac{7-\mathrm{x}}{3}\) to determine what values x can actually take. This approach makes it impossible to determine which statements must always be true.

Errors while executing the approach

1. Sign error when multiplying inequality by negative number
When solving \(5 < -\mathrm{x}\), students must multiply both sides by -1 to isolate x. A very common error is forgetting to flip the inequality sign, writing \(-5 < \mathrm{x}\) instead of the correct \(-5 > \mathrm{x}\) (or \(\mathrm{x} < -5\)). This fundamental mistake changes the entire constraint and leads to wrong conclusions about all three statements.

2. Incorrect evaluation of absolute value expression
For statement II, \(|\mathrm{x}+3| > 2\), students often struggle with the absolute value when \(\mathrm{x} < -5\). They might incorrectly conclude that since x+3 is negative, \(|\mathrm{x}+3| = \mathrm{x}+3\), rather than the correct \(|\mathrm{x}+3| = -(\mathrm{x}+3)\). This leads to wrong calculations about whether the absolute value is greater than 2.

3. Confusion with double negative in statement III
Statement III asks whether \(-(\mathrm{x}+5)\) is positive. Students often get confused by the negative sign in front of (x+5) and might incorrectly conclude that since x+5 is negative, \(-(\mathrm{x}+5)\) is also negative, rather than recognizing that the negative of a negative number is positive.

Errors while selecting the answer

1. Selecting partial answers when multiple statements are true
After correctly determining that both statements II and III are true, students might still select "B) III only" or "A) II only" because they focus on just one statement they're confident about, rather than recognizing that when multiple statements are true, they need to select the answer choice that includes ALL the true statements.