If x - 5 ≤ 2x - 3 and 2x - 1/2 ≥ 5x + 2 hold simultaneously, then which of the following values can x take?

Introduction / Context:
This aptitude question tests the ability to solve and interpret simultaneous linear inequalities in one variable. Such inequalities frequently appear in school mathematics, competitive exams, and real life situations where a quantity has to satisfy more than one condition at the same time.

Given Data / Assumptions:

• x is a real number.
• Inequality 1: x - 5 ≤ 2x - 3
• Inequality 2: 2x - 1/2 ≥ 5x + 2
• We must find which option satisfies both inequalities simultaneously.

Concept / Approach:
To solve a system of inequalities in a single variable, we solve each inequality separately to obtain an interval for x. Then we intersect the intervals to get the common solution set. Finally, we check which option values lie in that common interval.

Step-by-Step Solution:
From x - 5 ≤ 2x - 3, subtract x from both sides to get -5 ≤ x - 3. Add 3 on both sides: -2 ≤ x, so x ≥ -2. Now consider 2x - 1/2 ≥ 5x + 2. Subtract 2x: -1/2 ≥ 3x + 2. Subtract 2 on both sides: -1/2 - 2 ≥ 3x, which gives -5/2 ≥ 3x. Divide by 3 (positive), so x ≤ -5/6. Combined solution: -2 ≤ x ≤ -5/6. Check the options: -1 lies in the interval, 1 and 3 are greater than -5/6, and -3 is less than -2, while 0 is also greater than -5/6. So only x = -1 satisfies both inequalities.

Verification / Alternative check:
Substitute x = -1 into both inequalities: First: -1 - 5 = -6 and 2(-1) - 3 = -5, so -6 ≤ -5 is true. Second: 2(-1) - 1/2 = -2.5 and 5(-1) + 2 = -3, so -2.5 ≥ -3 is also true. Thus x = -1 is correct.

Why Other Options Are Wrong:
x = 1 gives 1 - 5 ≤ 2 - 3 which simplifies to -4 ≤ -1 (true), but 2(1) - 1/2 ≥ 5(1) + 2 gives 1.5 ≥ 7 which is false. x = 3 satisfies neither inequality simultaneously. x = -3 fails the first inequality because -3 - 5 ≤ -6 - 3 becomes -8 ≤ -9 which is false. x = 0 gives 0 ≥ -2 but fails the second inequality because -0.5 ≥ 2 is false.

Common Pitfalls:
Students often forget to find the intersection of solution sets and may stop after solving only one inequality. Another frequent mistake is mishandling signs when moving terms across the inequality, which changes the direction only when multiplying or dividing by a negative number, not by a positive number like 3 here.

Final Answer:
The only value of x that satisfies both inequalities is -1.