Solve the following system of linear inequalities in one variable: 1) 2x + 5 > 2 + 3x 2) 2x − 3 <= 4x − 5 From the options given, which value of x satisfies both inequalities simultaneously?

Introduction / Context:
This question asks you to solve a pair of linear inequalities in a single variable x and then select from several candidate values the one that satisfies both inequalities. Working with inequalities is a core topic in algebra and aptitude tests. The main ideas are very similar to solving equations, but one must carefully maintain the direction of the inequality and understand how solution sets intersect.

Given Data / Assumptions:

• First inequality: 2x + 5 > 2 + 3x.
• Second inequality: 2x − 3 <= 4x − 5.
• x is a real number, but we are given discrete options.
• We must find which option satisfies both inequalities at the same time.

Concept / Approach:
We solve each inequality separately to get its solution set in the form of an interval. The solution to the system is the intersection of these two intervals, that is, the set of all x values that satisfy both inequalities. Once the intersection is found, we check which of the given options lies within it. Since all coefficients in the inequalities are positive, there is no need to reverse the inequality sign when dividing by those coefficients.

Step-by-Step Solution:
1) Solve the first inequality: 2x + 5 > 2 + 3x. 2) Subtract 2x from both sides: 5 > 2 + x. 3) Subtract 2 from both sides: 3 > x, which means x < 3. 4) Now solve the second inequality: 2x − 3 <= 4x − 5. 5) Subtract 2x from both sides: −3 <= 2x − 5. 6) Add 5 to both sides: 2 <= 2x. 7) Divide both sides by 2 (positive): 1 <= x. 8) The second inequality therefore gives x >= 1. 9) Combine the two results: from the first inequality x < 3, and from the second x >= 1. 10) The solution set is 1 <= x < 3.

Verification / Alternative check:
Now test each option against the solution interval 1 <= x < 3. Option c is x = 2, which lies between 1 and 3 and should satisfy both inequalities. Check directly: for x = 2, the first inequality gives 2 * 2 + 5 = 9 and 2 + 3 * 2 = 8, so 9 > 8 is true. The second inequality gives 2 * 2 − 3 = 1 and 4 * 2 − 5 = 3, so 1 <= 3 is also true. Thus x = 2 satisfies both inequalities. The other options either fall outside the interval or fail one of the inequalities upon substitution.

Why Other Options Are Wrong:
Option a (−4) and option b (−2) are less than 1, so they do not satisfy the second inequality. Option d (4) is greater than or equal to 3, so it violates the first inequality. Option e (0) is less than 1, so again it fails the second inequality. Only option c (2) lies in the intersection 1 <= x < 3 and thus satisfies both inequalities.

Common Pitfalls:
Students sometimes combine inequalities incorrectly, for example by adding them together instead of intersecting their solution sets. Another common mistake is to mis move terms across the inequality sign, especially when negative coefficients are present, which would require reversing the inequality direction. In this problem, all divisions are by positive numbers, so the inequality directions stay the same. Writing each step clearly and then drawing a quick number line can help visualise the final solution interval.

Final Answer:
The value of x among the options that satisfies both inequalities is 2.