Solve the system of inequalities 7 + 3x ≥ 5 − x/2 and 2x + 3 ≤ 5 − 2x; which of the following values of x satisfies both conditions simultaneously?

Introduction / Context:
This problem involves solving two linear inequalities in one variable and then finding which option satisfies both at the same time. Such questions are common in aptitude tests and help you understand how to work with inequality symbols, manipulate expressions, and interpret ranges of possible solutions.

Given Data / Assumptions:

• First inequality: 7 + 3x ≥ 5 − x/2.
• Second inequality: 2x + 3 ≤ 5 − 2x.
• We need a value of x that satisfies both inequalities.

Concept / Approach:
Key ideas:

• Solve each inequality separately, obtaining a range of x values.
• Combine the two solution sets using intersection to see which x values work for both.
• Check each option against the final allowed interval.

Step-by-Step Solution:
First inequality: 7 + 3x ≥ 5 − x/2. Bring all x terms to one side: 3x + x/2 ≥ 5 − 7 = −2. 3x = 6x/2, so 6x/2 + x/2 = 7x/2. Thus 7x/2 ≥ −2, so 7x ≥ −4, giving x ≥ −4/7. Second inequality: 2x + 3 ≤ 5 − 2x. Bring x terms together: 2x + 2x ≤ 5 − 3 = 2. This gives 4x ≤ 2, so x ≤ 1/2. Combined solution: −4/7 ≤ x ≤ 1/2.

Verification / Alternative check:
Now test each option. Among the given values, 0 lies between −4/7 and 1/2, whereas 1 and 2 are greater than 1/2 and −1 and −2 are less than −4/7. Therefore x = 0 is the only value that satisfies both inequalities. Substituting x = 0 explicitly into both inequalities confirms that each side of both conditions is satisfied.

Why Other Options Are Wrong:
Option b: 1 violates x ≤ 1/2 because 1 is greater than 0.5.
Option c: 2 also violates the second inequality and is far outside the allowed range.
Option d: −1 is less than −4/7 and fails the first inequality.
Option e: −2 is even smaller and does not satisfy x ≥ −4/7.

Common Pitfalls:
Mistakes often occur when handling fractions, especially with the term −x/2, or when flipping the inequality sign incorrectly after multiplying or dividing by negative numbers. In this problem, all multipliers were positive, so the direction of the inequality does not change. Carefully isolating x in each inequality prevents confusion.

Final Answer:
The value of x that satisfies both inequalities is 0.