Solve the compound inequality system: 5x + 5 > 2 + 2x and 5x + 3 ≤ 4x + 5 Based on the solution set, which of the following given values can x take?

Introduction / Context:
This question tests solving two linear inequalities simultaneously and then checking which listed value satisfies both. When inequalities are connected by “and,” the solution must lie in the intersection of the two individual solution sets. After obtaining the valid interval for x, the final step is to test the answer options against that interval. Careful handling of inequality signs is important, especially when subtracting or dividing by negative numbers (though this problem only requires subtracting and adding).

Given Data / Assumptions:

• Inequality 1: 5x + 5 > 2 + 2x • Inequality 2: 5x + 3 ≤ 4x + 5 • Both must hold at the same time (logical AND) • Required: choose an option value that satisfies both

Concept / Approach:
Solve each inequality separately: 1) Simplify to get x > some value. 2) Simplify to get x ≤ some value. Then intersect the results to form one combined interval. Finally test the given options and find which lie in that interval.

Step-by-Step Solution:
1) Solve inequality 1: 5x + 5 > 2 + 2x 5x - 2x > 2 - 5 3x > -3 x > -1 2) Solve inequality 2: 5x + 3 ≤ 4x + 5 5x - 4x ≤ 5 - 3 x ≤ 2 3) Combine (AND means intersection): -1 < x ≤ 2 4) Test options: x = 1 lies between -1 and 2 (and equals neither forbidden boundary), so it works.

Verification / Alternative check:
Direct substitution for x = 1: Inequality 1: 5(1)+5 > 2+2(1) → 10 > 4 (true). Inequality 2: 5(1)+3 ≤ 4(1)+5 → 8 ≤ 9 (true). So x = 1 satisfies both conditions. Options like 3 fail because they exceed 2, and -2 or -3 fail because they are not greater than -1.

Why Other Options Are Wrong:
• 3: violates x ≤ 2. • -2 and -3: violate x > -1. • 0 is also valid for the interval, but it is not listed as the intended correct option in this set; here the correct listed value satisfying both is 1.

Common Pitfalls:
• Treating “and” as “or” and taking a union instead of an intersection. • Mishandling strict (>) versus non-strict (≤) inequality boundaries.

Final Answer:
1