Consider the fraction with numerator 6x + 1 and denominator 7 − 4x. For x in the closed interval [-2, 2], determine all values of x for which the numerator is greater than the denominator (i.e., 6x + 1 > 7 − 4x).

Introduction: We must compare the linear expressions 6x + 1 and 7 − 4x over the interval [-2, 2]. The task reduces to solving a first-degree inequality and then intersecting with the given domain.

Given Data / Assumptions:

• Inequality: 6x + 1 > 7 − 4x.
• Domain: −2 ≤ x ≤ 2.
• Strict inequality is required (“greater than”).

Concept / Approach: Rearrange to gather x-terms on one side and constants on the other. Solve for x and then apply the domain restriction to express the final answer in interval language.

Step-by-Step Solution:

Verification / Alternative check: Test a point: x = 1 (allowed). 6(1)+1 = 7; 7 − 4(1) = 3 → 7 > 3 holds. Test x = 0.5 (just below 3/5): 6(0.5)+1 = 4; 7 − 4(0.5) = 5 → 4 > 5 is false; boundary logic confirmed.

Why Other Options Are Wrong: Options including “≤” at 3/5 treat the strict inequality incorrectly; other ranges either over- or under-include values outside the solution set.

Common Pitfalls: Using ≥ instead of > at the boundary or forgetting to intersect with the domain [-2, 2].

Final Answer: 3/5 < x ≤ 2