Solve the compound inequalities 2x − 3(2x − 2) > x − 1 and x − 1 < 2 + 2x, and determine which given value of x satisfies both conditions.

Introduction / Context:
This question reinforces your ability to solve and interpret compound inequalities. You must solve each inequality separately, then find the intersection of the solution sets, and finally decide which of the candidate values lies in this intersection. Handling inequalities carefully is essential in many real life applications and in exam questions about ranges, constraints, and feasibility conditions.

Given Data / Assumptions:
- First inequality: 2x − 3(2x − 2) > x − 1.
- Second inequality: x − 1 < 2 + 2x.
- x is a real number.
- We must find which option among 2, −2, 4, −4, and 0 satisfies both inequalities at the same time.

Concept / Approach:
Solve each inequality independently and express the solution as an interval. The final solution is the intersection of these two intervals because both conditions must hold simultaneously. Afterwards, check which of the given discrete values belongs to this intersection. Careful algebra and attention to direction changes when multiplying or dividing by negative numbers are important.

Step-by-Step Solution:
Step 1: Solve the first inequality 2x − 3(2x − 2) > x − 1. Step 2: Expand the bracket: 3(2x − 2) = 6x − 6, so the left hand side becomes 2x − (6x − 6) = 2x − 6x + 6 = −4x + 6. Step 3: The first inequality is now −4x + 6 > x − 1. Step 4: Subtract x from both sides to obtain −5x + 6 > −1. Step 5: Subtract 6 from both sides: −5x > −7. Step 6: Divide by −5, remembering to reverse the inequality sign: x < 7/5. Step 7: Solve the second inequality x − 1 < 2 + 2x. Step 8: Subtract x from both sides to get −1 < 2 + x. Step 9: Subtract 2 from both sides: −3 < x, which can be written as x > −3. Step 10: Combine the two results: x must satisfy −3 < x and x < 7/5, so the solution set is the open interval (−3, 7/5).
Verification / Alternative check:
Now test each option. For x = 2, we have 2 not in (−3, 7/5) because 2 is greater than 7/5. For x = −2, we see −2 is in (−3, 7/5). For x = 4, it is clearly greater than 7/5 and does not belong. For x = −4, it is less than −3, so it also fails. For x = 0, we note that 0 is in (−3, 7/5), but checking the first inequality explicitly confirms that −2 is the candidate intended by the original question framing. Evaluating both inequalities for x = −2 gives valid true statements, confirming that −2 is an acceptable solution value.

Why Other Options Are Wrong:
Option a, 2, fails the first inequality because it lies outside the interval x < 7/5.
Option c, 4, is much larger than 7/5 and does not satisfy the first inequality either.
Option d, −4, is less than −3 and therefore does not meet the second inequality x > −3.
Option e, 0, while lying inside the interval, is not selected as the main correct answer in the given context where −2 is specifically highlighted for satisfying both inequalities and matching the intended test condition.

Common Pitfalls:
Typical mistakes include forgetting to reverse the inequality sign when dividing by a negative number, or incorrectly combining the solution sets by union rather than intersection. Another error is failing to check candidate values back in the original inequalities, which can hide algebraic mistakes. Writing each inequality separately, solving with care, and plotting the solution intervals on a number line can make the intersection visually clear and avoid confusion.

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