If the inequalities 2x + 2(4 + 3x) < 2 + 3x and 2 + 3x > 2x + x/2 both hold simultaneously, determine which given value of x satisfies both conditions.

Introduction / Context:
This question involves solving two linear inequalities in one variable and then finding a value of x from the given options that satisfies both simultaneously. Inequalities are a core algebra topic in aptitude tests, and combining several inequality conditions is a common way to check deeper understanding. Careful step by step manipulation is needed to avoid sign errors.

Given Data / Assumptions:

• Inequality 1: 2x + 2(4 + 3x) < 2 + 3x.

• Inequality 2: 2 + 3x > 2x + x/2.

• x is a real number, and we must select the correct value from the options −3, 1, 0, −1 and −5.

Concept / Approach:
We simplify each inequality separately to obtain an interval of allowed x values. Inequality 1 will produce one condition (such as x less than some number), and inequality 2 will produce another condition (for example x greater than some number). The values of x that satisfy both conditions must lie in the intersection of the two intervals. Finally, we test each option to see which falls in this intersection.

Step-by-Step Solution:
Step 1: Simplify Inequality 1: 2x + 2(4 + 3x) < 2 + 3x. Step 2: Expand: 2x + 8 + 6x < 2 + 3x, so 8x + 8 < 2 + 3x. Step 3: Subtract 3x from both sides: 5x + 8 < 2. Step 4: Subtract 8: 5x < −6, hence x < −6/5 = −1.2. Step 5: Simplify Inequality 2: 2 + 3x > 2x + x/2. Step 6: Combine terms on the right: 2x + x/2 = (4x/2 + x/2) = 5x/2. Step 7: So 2 + 3x > 5x/2. Multiply every term by 2 to clear the denominator: 4 + 6x > 5x. Step 8: Subtract 5x: 4 + x > 0, so x > −4. Step 9: Combine both conditions: from Inequality 1, x < −1.2 and from Inequality 2, x > −4. Therefore −4 < x < −1.2.

Verification / Alternative check:
We now test the options. x = −3 lies between −4 and −1.2, so it is in the allowed interval. Plugging x = −3 into Inequality 1 gives 2(−3) + 2(4 + 3(−3)) = −6 + 2(4 − 9) = −6 + 2(−5) = −16. The right side is 2 + 3(−3) = 2 − 9 = −7. Since −16 < −7, Inequality 1 holds. For Inequality 2, 2 + 3(−3) = −7 and 2(−3) + (−3)/2 = −6 − 1.5 = −7.5, and −7 > −7.5, so Inequality 2 also holds. None of the other options fall strictly between −4 and −1.2, so they cannot satisfy both inequalities together.

Why Other Options Are Wrong:
x = 1 and x = 0 are greater than −1.2, so they fail Inequality 1 which requires x < −1.2. x = −1 is greater than −1.2 and also fails Inequality 1. x = −5 is less than −4 and fails Inequality 2, which requires x > −4. Only x = −3 lies in the intersection of both solution intervals and satisfies both inequalities when substituted explicitly.

Common Pitfalls:
Students sometimes make sign errors when moving terms across the inequality or forget to reverse the inequality sign when multiplying by a negative number. In this problem no sign reversal is needed, but careful handling of each arithmetic step is still essential. Another pitfall is to stop after solving only one inequality without checking the second, which can lead to incorrect conclusions about which options are valid.

Final Answer:
The value of x that satisfies both inequalities is −3.